A topological space is what remains when we stop measuring distances but still want to speak about nearness, convergence, continuity, and shape. The central problem is that many mathematical objects behave continuously without carrying a preferred metric: quotient spaces, function spaces, algebraic varieties, spectra of rings, and manifolds assembled from charts. Topology gives a language for continuity that depends only on which subsets count as open. The first warning comes from identifying points. Start with the interval $[0,1]$ and glue $0$ to $1$. If we keep the usual distance inherited from the line, the endpoints remain far apart, even though the quotient construction asks them to become the same point. The right structure is not the old metric; it is the collection of subsets whose preimages in $[0,1]$ are open. This produces the circle topology and shows why openness, not distance, is the primitive notion. [example: Gluing an Interval into a Circle] Let $S^1=\{x\in\mathbb R^2:|x|=1\}$, and define \begin{align*} q:[0,1] &\to S^1 \\ t &\mapsto (\cos(2\pi t),\sin(2\pi t)). \end{align*} First, \begin{align*} q(0)&=(\cos 0,\sin 0)=(1,0),\\ q(1)&=(\cos 2\pi,\sin 2\pi)=(1,0), \end{align*} so the quotient map identifies the two endpoints. Fix $0<\varepsilon<\frac12$, and let \begin{align*} V_\varepsilon &=q\big([0,\varepsilon)\cup(1-\varepsilon,1]\big)\subset S^1. \end{align*} We compute its preimage. If $q(t)=q(s)$ for $s,t\in[0,1]$, then \begin{align*} e^{2\pi i t}&=e^{2\pi i s},\\ e^{2\pi i(t-s)}&=1, \end{align*} so $t-s\in\mathbb Z$. Since $t-s\in[-1,1]$, this gives $t=s$, or $(t,s)=(1,0)$, or $(t,s)=(0,1)$. Therefore no point of $[\varepsilon,1-\varepsilon]$ maps into $V_\varepsilon$, and \begin{align*} q^{-1}(V_\varepsilon) &=[0,\varepsilon)\cup(1-\varepsilon,1]. \end{align*} This set is open in $[0,1]$ with the [subspace topology](/page/Subspace%20Topology) because \begin{align*} [0,\varepsilon)\cup(1-\varepsilon,1] &=[0,1]\cap\big((-\varepsilon,\varepsilon)\cup(1-\varepsilon,1+\varepsilon)\big), \end{align*} and $(-\varepsilon,\varepsilon)\cup(1-\varepsilon,1+\varepsilon)$ is open in $\mathbb R$. Thus an arc around $(1,0)$ is locally one connected piece on the circle, even though its preimage appears as two pieces at opposite ends of the interval. [/example] The lesson of the example is that the topology on a space records which tests for local membership are allowed. Once this is known, continuity becomes a statement about inverse images of open sets, convergence can be phrased through neighbourhoods, and constructions can be defined by universal properties rather than by formulas for distances. ## Definition The goal is to isolate the smallest amount of structure needed to decide which subsets are open. The axioms should include arbitrary unions because local conditions may vary from point to point, and finite intersections because satisfying finitely many local conditions at once should remain local. They should not include arbitrary intersections: intersecting infinitely many open intervals around $0$ in $\mathbb R$ can collapse to the single point $\{0\}$, which should not be open in the usual topology. [definition: Topological Space] A topological space is a pair $(X, \tau)$ where $X$ is a set and $\tau \subset \mathcal P(X)$ is a collection of subsets of $X$ such that: \begin{align*} \varnothing &\in \tau, \\ X &\in \tau, \end{align*} for every family $(U_i)_{i \in I}$ with $U_i \in \tau$ for all $i \in I$, \begin{align*} \bigcup_{i \in I} U_i &\in \tau, \end{align*} and for every finite family $U_1,\ldots,U_n \in \tau$, \begin{align*} \bigcap_{i=1}^n U_i &\in \tau. \end{align*} The elements of $\tau$ are called open sets. [/definition] ## Basic Open and Closed Constructions ### Closed and Subspace Sets After open sets have been chosen, the next problem is how to describe sets determined by exclusion rather than by local entry. Limit points, compactness arguments, and boundary phenomena are usually controlled by sets whose complements are open, so we need the dual vocabulary. [definition: Closed Set] Let $(X,\tau)$ be a topological space. A subset $F \subset X$ is closed if $X \setminus F \in \tau$. [/definition] The definition by complements is useful because it converts the permanence rules for open sets into dual permanence rules for closed sets. Without a separate permanence statement, every later argument about closures, boundaries, and compact subspaces would have to repeatedly translate through complements. The next theorem packages that translation into a tool we can use whenever closed conditions are imposed simultaneously or finitely many closed obstructions are combined. [quotetheorem:4951] These closure rules make closed conditions usable in practice: arbitrary simultaneous closed constraints remain closed, and finitely many closed obstructions can be combined without leaving the category of closed sets. This raises the construction problem for subsets: if a space sits inside a larger one, which of the ambient open sets should still count as open after restriction? [definition: Subspace Topology] Let $(X,\tau)$ be a topological space and let $A \subset X$. The subspace topology on $A$ is \begin{align*} \tau_A &= \{A \cap U : U \in \tau\}. \end{align*} The pair $(A,\tau_A)$ is called a subspace of $(X,\tau)$. [/definition] The subspace topology explains why $[0,\varepsilon)$ can be open in $[0,1]$ even though it is not open in $\mathbb R$. The point of the next example is to make that dependence on the ambient space concrete. [example: Open in a Subspace but Not in the Ambient Space] In $\mathbb R$ with its usual topology, give $[0,1]$ the subspace topology. We show that $[0,1/2)$ is open in the subspace $[0,1]$ but not open in the ambient space $\mathbb R$. First, \begin{align*} [0,1]\cap(-1/2,1/2) &=\{x\in\mathbb R:0\le x\le 1\text{ and }-1/2<x<1/2\} \\ &=\{x\in\mathbb R:0\le x<1/2\} \\ &=[0,1/2). \end{align*} Since $(-1/2,1/2)$ is open in $\mathbb R$, the equality above shows that $[0,1/2)$ is open in $[0,1]$ by the definition of the subspace topology. It is not open in $\mathbb R$. Indeed, if $[0,1/2)$ were open in $\mathbb R$, then because $0\in[0,1/2)$ there would be some $\varepsilon>0$ such that \begin{align*} (-\varepsilon,\varepsilon)\subset[0,1/2). \end{align*} But $-\varepsilon/2\in(-\varepsilon,\varepsilon)$, while $-\varepsilon/2<0$, so \begin{align*} -\varepsilon/2\notin[0,1/2). \end{align*} This contradicts the containment. Thus openness depends on the ambient topology: $[0,1/2)$ is open inside $[0,1]$, but not inside $\mathbb R$. [/example] ### Extreme Topologies At the two extremes, topology can either remember every subset or almost none of them. What happens if every function from $X$ should be continuous? This forces every subset of $X$ to pass the open-set test, giving the finest possible topology. [definition: Discrete Topology] Let $X$ be a set. The discrete topology on $X$ is \begin{align*} \tau &= \mathcal P(X). \end{align*} [/definition] In the discrete topology, every subset is open. That extreme choice should have a visible consequence for maps: once continuity is defined by preimages of open sets, a discrete domain will have already made every possible preimage legal. The opposite problem asks how little topology a set can carry while still satisfying the axioms. [definition: Indiscrete Topology] Let $X$ be a set. The indiscrete topology on $X$ is \begin{align*} \tau &= \{\varnothing, X\}. \end{align*} [/definition] The indiscrete topology is the coarsest topology on $X$. It creates a useful failure mode: if open sets cannot distinguish points, then topology cannot yet distinguish local behavior either. Later, after neighbourhoods and convergence have been defined, this same example explains why [uniqueness of limits](/theorems/625) requires an additional separation condition. ## Neighbourhoods, Closure, and Interior Open sets describe the global topology, but local arguments need the vocabulary of what happens near a point. A neighbourhood is a set large enough to contain an open region around the point; it allows arguments to use sets that are not themselves open while preserving a local open core. [definition: Neighbourhood] Let $(X,\tau)$ be a topological space and let $x \in X$. A subset $N \subset X$ is a neighbourhood of $x$ if there exists $U \in \tau$ such that \begin{align*} x \in U \subset N. \end{align*} [/definition] Neighbourhoods turn topological statements into pointwise tests. The next question is which points of a set have genuine local room inside the set, rather than lying at its edge. [definition: Interior] Let $(X,\tau)$ be a topological space and let $A \subset X$. The interior of $A$ is \begin{align*} A^\circ &= \bigcup\{U \in \tau : U \subset A\}. \end{align*} [/definition] The definition builds the interior from all open sets already lying inside $A$. The following maximality statement is the reason this construction behaves like the safe open part of $A$. [quotetheorem:4954] This maximality is why interior is the right replacement for "the open part of $A$". The complementary problem is what to do with points that may not belong to $A$ but cannot be separated from it by any open test. [definition: Closure] Let $(X,\tau)$ be a topological space and let $A \subset X$. The closure of $A$ is \begin{align*} \overline{A} &= \bigcap\{F \subset X : F \text{ is closed and } A \subset F\}. \end{align*} [/definition] The definition builds the closure from all closed sets that already contain $A$. The following minimality statement is what lets closure replace many separate closed-containment arguments. [quotetheorem:4956] The closure is therefore the smallest closed environment in which $A$ can live. Once interior and closure are available, the remaining interface question is where the safe inside of $A$ ends and unavoidable contact with $A$ begins. [definition: Boundary] Let $(X,\tau)$ be a topological space and let $A \subset X$. The boundary of $A$ is \begin{align*} \partial A &= \overline{A} \setminus A^\circ. \end{align*} [/definition] Boundary points are those at which neighbourhood tests cannot treat the set as purely inside. The next example shows why dense subsets can have empty interior but large closure. [example: Closure and Boundary in the Real Line] In $\mathbb R$ with its usual topology, let $A=(0,1)\cap\mathbb Q$. We compute the interior, closure, and boundary: \begin{align*} A^\circ &= \varnothing, \\ \overline{A} &= [0,1], \\ \partial A &= [0,1]. \end{align*} First suppose $x\in A^\circ$. Then there is an open interval $(x-\delta,x+\delta)$ with $\delta>0$ such that \begin{align*} x\in (x-\delta,x+\delta)\subset A. \end{align*} Since $x\in A$, the number $x$ is rational. Choose $n\in\mathbb N$ with \begin{align*} \frac{\sqrt 2}{n}<\delta. \end{align*} Then \begin{align*} x< x+\frac{\sqrt 2}{n}<x+\delta, \end{align*} so $x+\frac{\sqrt 2}{n}\in (x-\delta,x+\delta)$. But $x+\frac{\sqrt 2}{n}$ is irrational, because otherwise \begin{align*} \frac{\sqrt 2}{n} &=\left(x+\frac{\sqrt 2}{n}\right)-x \end{align*} would be rational, forcing $\sqrt 2$ to be rational. This contradicts $(x-\delta,x+\delta)\subset A\subset\mathbb Q$. Hence $A^\circ=\varnothing$. For the closure, first $A\subset[0,1]$, and $[0,1]$ is closed in $\mathbb R$, so \begin{align*} \overline A\subset[0,1]. \end{align*} Now let $x\in[0,1]$, and let $(x-\delta,x+\delta)$ be any open interval around $x$. If $x<1$, choose $n\in\mathbb N$ so large that \begin{align*} \frac1n<\min(\delta,1-x), \end{align*} and set $m=\lfloor nx\rfloor+1$. Then $m>nx$ and $m\le nx+1$, so $r=m/n$ is rational and satisfies \begin{align*} x<r\le x+\frac1n<x+\delta. \end{align*} Also $r<1$ because $x+\frac1n<1$. If $x=1$, choose $n\in\mathbb N$ with $\frac1n<\delta$; then the rational number \begin{align*} r=1-\frac1n \end{align*} satisfies \begin{align*} 1-\delta<1-\frac1n<1. \end{align*} Thus every open interval around every $x\in[0,1]$ meets $A$, so $[0,1]\subset\overline A$. Therefore $\overline A=[0,1]$. Finally, by the definition of boundary, \begin{align*} \partial A &=\overline A\setminus A^\circ \\ &=[0,1]\setminus\varnothing \\ &=[0,1]. \end{align*} Thus $A$ has no open part in $\mathbb R$, but its closure fills the whole closed interval from $0$ to $1$. [/example] The example suggests that interior and closure are not independent operations. Since both are controlled by complements of open or closed sets, we need a precise duality that lets arguments move between the two languages. [quotetheorem:1014] The theorem is the practical bridge between open-set and closed-set methods. It lets an argument move to whichever side has the better permanence property: interior language is often natural for local openness, while closure language is often natural for limiting and containment arguments. ## Bases and Generating Topologies ### Bases Listing every [open set](/page/Open%20Set) is usually inefficient. On $\mathbb R$, it is better to specify open intervals and allow arbitrary unions. On manifolds, coordinate balls play a similar role. A basis is a small supply of open sets from which the whole topology can be reconstructed. [definition: Basis for a Topology] Let $X$ be a set. A collection $\mathcal B \subset \mathcal P(X)$ is a basis for a topology on $X$ if: \begin{align*} X &= \bigcup_{B \in \mathcal B} B, \end{align*} and whenever $B_1,B_2 \in \mathcal B$ and $x \in B_1 \cap B_2$, there exists $B_3 \in \mathcal B$ such that \begin{align*} x \in B_3 \subset B_1 \cap B_2. \end{align*} [/definition] The second basis axiom says that finite intersections can be refined locally by basic open sets. To use a basis as data, we need to say exactly which subsets become open after the basis is declared. [definition: Topology Generated by a Basis] Let $X$ be a set and let $\mathcal B$ be a basis for a topology on $X$. The topology generated by $\mathcal B$ is \begin{align*} \tau_{\mathcal B} &= \{U \subset X : \text{for every } x \in U, \text{ there exists } B \in \mathcal B \text{ with } x \in B \subset U\}. \end{align*} [/definition] A local test is useful only if it actually produces a topology; otherwise a proposed basis would not define a space. The theorem below resolves that existence problem and shows that no hidden open sets are being added beyond unions of basic opens. [quotetheorem:4959] The theorem justifies the usual practice of defining a topology by giving basic opens. Once a basis has been checked, the actual open sets are obtained by freely taking unions of those local building blocks. The standard model is the usual topology on the real line. [example: Open Intervals Generate the Usual Topology] Let \begin{align*} \mathcal B &= \{(a,b) \subset \mathbb R : a,b \in \mathbb R,\ a<b\}. \end{align*} We first check that $\mathcal B$ is a basis. If $x\in\mathbb R$, then \begin{align*} x-1<x<x+1, \end{align*} so $x\in(x-1,x+1)\in\mathcal B$. Hence \begin{align*} \mathbb R=\bigcup_{B\in\mathcal B}B. \end{align*} Now let $(a,b),(c,d)\in\mathcal B$ and suppose $x\in(a,b)\cap(c,d)$. Then \begin{align*} a<x<b,\qquad c<x<d. \end{align*} Thus \begin{align*} \max\{a,c\}<x<\min\{b,d\}. \end{align*} Set \begin{align*} B_x&=(\max\{a,c\},\min\{b,d\}). \end{align*} Then $B_x\in\mathcal B$, $x\in B_x$, and if $y\in B_x$, then \begin{align*} \max\{a,c\}<y<\min\{b,d\}. \end{align*} Since $a\le \max\{a,c\}$ and $\min\{b,d\}\le b$, we get \begin{align*} a<y<b, \end{align*} so $y\in(a,b)$. Similarly, since $c\le\max\{a,c\}$ and $\min\{b,d\}\le d$, we get \begin{align*} c<y<d, \end{align*} so $y\in(c,d)$. Therefore \begin{align*} B_x\subset (a,b)\cap(c,d). \end{align*} The topology generated by $\mathcal B$ is the usual topology on $\mathbb R$. Indeed, if $U$ is open in the usual topology and $x\in U$, then by the usual interval-neighbourhood condition there is $\varepsilon>0$ such that \begin{align*} (x-\varepsilon,x+\varepsilon)\subset U. \end{align*} Since $(x-\varepsilon,x+\varepsilon)\in\mathcal B$ and contains $x$, this shows $U$ is open in the topology generated by $\mathcal B$. Conversely, if $U$ is open in the topology generated by $\mathcal B$ and $x\in U$, then there exists $(a,b)\in\mathcal B$ such that \begin{align*} x\in(a,b)\subset U. \end{align*} From $a<x<b$, the positive number \begin{align*} \varepsilon&=\frac12\min\{x-a,b-x\} \end{align*} satisfies \begin{align*} 0<\varepsilon\le x-a,\qquad 0<\varepsilon\le b-x. \end{align*} Hence \begin{align*} a\le x-\varepsilon < x+\varepsilon \le b, \end{align*} and therefore \begin{align*} (x-\varepsilon,x+\varepsilon)\subset(a,b)\subset U. \end{align*} So $U$ is open in the usual topology. Thus the open intervals generate exactly the usual topology on $\mathbb R$. [/example] ### Subbases Sometimes a proposed generating family is not closed enough under intersections to be a basis, but it still should generate a topology after allowing finite intersections first. Product spaces create exactly this need: the natural tests constrain one coordinate at a time, not all coordinates at once. [definition: Subbasis for a Topology] Let $X$ be a set. A collection $\mathcal S \subset \mathcal P(X)$ is a subbasis for a topology on $X$ if \begin{align*} X &= \bigcup_{S \in \mathcal S} S. \end{align*} [/definition] A subbasis by itself is not yet a topology, because finite intersections of declared tests may be missing. If two subbasic conditions must hold at once, their intersection should count as a basic local test; without adding these finite intersections, arbitrary unions would not have a reliable local shape. This raises the precise construction problem: given only the declared subbasic tests, what is the smallest topology forced by requiring those tests to be open? The answer is to form a basis from all finite simultaneous tests, then generate a topology from that basis. [definition: Topology Generated by a Subbasis] Let $X$ be a set and let $\mathcal S$ be a subbasis for a topology on $X$. The topology generated by $\mathcal S$ is the topology generated by the basis consisting of all finite intersections \begin{align*} S_1 \cap \cdots \cap S_n, \end{align*} where $n \in \mathbb N$ and $S_1,\ldots,S_n \in \mathcal S$. [/definition] Subbases are a controlled way to say that several tests are declared open and that the topology contains no extra open sets beyond those forced by the axioms. ## Continuity and Homeomorphism ### Continuous Maps The main reason to build topological spaces is to make continuity independent of coordinates and metrics. The familiar epsilon-delta definition on metric spaces becomes a preimage condition: a map is continuous if open tests in the target pull back to open tests in the source. [definition: Continuous Map] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A map \begin{align*} f: X &\to Y \end{align*} is continuous if for every $V \in \tau_Y$, \begin{align*} f^{-1}(V) &\in \tau_X. \end{align*} [/definition] This definition is contravariant because inverse images preserve arbitrary unions and finite intersections for every function. The discrete topology now has its promised universal property. If the domain is discrete, every preimage subset is open automatically, so maps out of a discrete space are the least restrictive continuous maps. [quotetheorem:4952] This result is useful because it isolates the role of the domain topology. A discrete domain imposes no continuity constraint on a function: all the topological work has already been done by declaring every subset open. The limitation is equally important: discreteness says nothing about maps into $X$, where the target topology controls which preimages must be checked. The next example shows why direct images cannot be used as the definition of continuity. [example: Direct Images Do Not Preserve Openness] Let \begin{align*} f:\mathbb R &\to \mathbb R \\ x &\mapsto x^2. \end{align*} The interval $(-1,1)$ is open in $\mathbb R$. Its image is \begin{align*} f((-1,1)) &=\{x^2:-1<x<1\}. \end{align*} If $-1<x<1$, then $0\le x^2<1$, so \begin{align*} f((-1,1))\subset [0,1). \end{align*} Conversely, if $y\in[0,1)$, then $0\le y<1$, so $0\le \sqrt y<1$ and \begin{align*} \sqrt y\in(-1,1),\qquad f(\sqrt y)=(\sqrt y)^2=y. \end{align*} Hence \begin{align*} [0,1)\subset f((-1,1)), \end{align*} and therefore \begin{align*} f((-1,1))=[0,1). \end{align*} The set $[0,1)$ is not open in $\mathbb R$. If it were open, then since $0\in[0,1)$ there would be some $\varepsilon>0$ such that \begin{align*} (-\varepsilon,\varepsilon)\subset[0,1). \end{align*} But $-\varepsilon/2\in(-\varepsilon,\varepsilon)$ and $-\varepsilon/2<0$, so \begin{align*} -\varepsilon/2\notin[0,1), \end{align*} contradicting the containment. The map $f$ is nevertheless continuous. For an open interval $(a,b)\subset\mathbb R$ with $a<b$, its preimage is open because \begin{align*} f^{-1}((a,b)) &=\{x\in\mathbb R:a<x^2<b\}. \end{align*} If $b\le 0$, then $x^2\ge 0$ for all $x$, so \begin{align*} f^{-1}((a,b))=\varnothing. \end{align*} If $a<0<b$, then $a<x^2$ holds for every $x$, and $x^2<b$ is equivalent to $-\sqrt b<x<\sqrt b$, so \begin{align*} f^{-1}((a,b))=(-\sqrt b,\sqrt b). \end{align*} If $0\le a<b$, then $a<x^2<b$ is equivalent to either $-\sqrt b<x<-\sqrt a$ or $\sqrt a<x<\sqrt b$, so \begin{align*} f^{-1}((a,b)) &=(-\sqrt b,-\sqrt a)\cup(\sqrt a,\sqrt b). \end{align*} In every case the preimage of an open interval is open, and arbitrary open subsets of $\mathbb R$ are unions of open intervals. Thus inverse images preserve openness here, while direct images need not preserve openness. [/example] Once continuity is defined by inverse images, we need it to be stable under applying one continuous process after another. Otherwise continuous change would not form a usable theory of maps between spaces. [quotetheorem:4960] This stability is what lets continuous maps be chained without rechecking the open-set test from scratch. It is also the reason homeomorphisms form the natural notion of sameness for topological spaces. ### Homeomorphisms Composition lets topological spaces form a category of structure-preserving maps. The next question is when two spaces represent the same topological object, so that each can be recovered continuously from the other. [definition: Homeomorphism] Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A map $f: X \to Y$ is a homeomorphism if $f$ is bijective, $f$ is continuous, and \begin{align*} f^{-1}: Y &\to X \end{align*} denotes the inverse function of $f$ and is continuous. [/definition] A homeomorphism identifies two spaces from the topological point of view. The inverse condition is essential, as the next example shows. [example: A Continuous Bijection That Is Not a Homeomorphism] Let $X=[0,1)$ with the subspace topology from $\mathbb R$, and define \begin{align*} f:X&\to S^1\\ t&\mapsto(\cos(2\pi t),\sin(2\pi t)). \end{align*} The map is continuous: the coordinate functions $t\mapsto\cos(2\pi t)$ and $t\mapsto\sin(2\pi t)$ are continuous real-valued functions on $X$, so for every open set $O\subset\mathbb R^2$, the set \begin{align*} f^{-1}(O) &=\{t\in X:(\cos(2\pi t),\sin(2\pi t))\in O\} \end{align*} is open in $X$. Since open sets in $S^1$ have the form $S^1\cap O$ with $O$ open in $\mathbb R^2$, we have \begin{align*} f^{-1}(S^1\cap O) &=f^{-1}(O), \end{align*} because $f(t)\in S^1$ for every $t\in X$. Hence the preimage of every open subset of $S^1$ is open in $X$. The map is injective. If $f(t)=f(s)$ for $s,t\in[0,1)$, then \begin{align*} \cos(2\pi t)&=\cos(2\pi s),\\ \sin(2\pi t)&=\sin(2\pi s), \end{align*} so \begin{align*} e^{2\pi i t}&=e^{2\pi i s},\\ e^{2\pi i(t-s)}&=1. \end{align*} Thus $t-s\in\mathbb Z$. But $s,t\in[0,1)$ imply \begin{align*} -1<t-s<1, \end{align*} so the only possible integer value is $t-s=0$. Therefore $t=s$. The map is surjective. If $(x,y)\in S^1$, then $x^2+y^2=1$, so there is an angle $\theta\in[0,2\pi)$ such that \begin{align*} x&=\cos\theta,\\ y&=\sin\theta. \end{align*} Setting \begin{align*} t&=\frac{\theta}{2\pi} \end{align*} gives $t\in[0,1)$ and \begin{align*} f(t) &=(\cos(2\pi t),\sin(2\pi t))\\ &=(\cos\theta,\sin\theta)\\ &=(x,y). \end{align*} Thus $f$ is a continuous bijection. It remains to show that $f$ is not a homeomorphism. Let \begin{align*} g:S^1&\to X \end{align*} be the inverse function. The set $[0,1/4)$ is open in $X$, because \begin{align*} [0,1/4) &=[0,1)\cap(-1/4,1/4), \end{align*} and $(-1/4,1/4)$ is open in $\mathbb R$. Its preimage under $g$ is \begin{align*} g^{-1}([0,1/4)) &=f([0,1/4))\\ &=\{(\cos(2\pi t),\sin(2\pi t)):0\le t<1/4\}. \end{align*} This set contains $(1,0)$, since $f(0)=(1,0)$. If $g^{-1}([0,1/4))$ were open in $S^1$, then there would be some $\delta>0$ such that \begin{align*} S^1\cap B_{\mathbb R^2}((1,0),\delta) &\subset g^{-1}([0,1/4)). \end{align*} Choose $n\in\mathbb N$ with \begin{align*} \frac{2\pi}{n}<\delta. \end{align*} Set \begin{align*} p_n&=f\left(1-\frac1n\right)\\ &=\left(\cos\left(2\pi-\frac{2\pi}{n}\right),\sin\left(2\pi-\frac{2\pi}{n}\right)\right). \end{align*} Then $p_n\in S^1$, and \begin{align*} \|p_n-(1,0)\|^2 &=\left(\cos\left(2\pi-\frac{2\pi}{n}\right)-1\right)^2 +\sin^2\left(2\pi-\frac{2\pi}{n}\right)\\ &=\cos^2\left(2\pi-\frac{2\pi}{n}\right) -2\cos\left(2\pi-\frac{2\pi}{n}\right)+1 +\sin^2\left(2\pi-\frac{2\pi}{n}\right)\\ &=2-2\cos\left(2\pi-\frac{2\pi}{n}\right). \end{align*} Using $1-\cos u\le u^2/2$ with $u=2\pi/n$, we get \begin{align*} \|p_n-(1,0)\|^2 &\le \left(\frac{2\pi}{n}\right)^2, \end{align*} and therefore \begin{align*} \|p_n-(1,0)\| &\le \frac{2\pi}{n} <\delta. \end{align*} So $p_n\in S^1\cap B_{\mathbb R^2}((1,0),\delta)$. However, \begin{align*} g(p_n) &=g\left(f\left(1-\frac1n\right)\right)\\ &=1-\frac1n. \end{align*} For $n\ge2$, \begin{align*} 1-\frac1n\ge\frac12>\frac14, \end{align*} so \begin{align*} g(p_n)\notin[0,1/4). \end{align*} Thus \begin{align*} p_n\notin g^{-1}([0,1/4)), \end{align*} contradicting the assumed containment. Hence $g^{-1}([0,1/4))$ is not open in $S^1$, so $g$ is not continuous. Therefore $f$ is a continuous bijection but not a homeomorphism: the topology on $[0,1)$ does not glue the missing endpoint back to $0$. [/example] This example is a central warning: topology is not preserved by bijection alone. The inverse map must respect the same open-set structure. ## Separation and Countability ### Separation Axioms The indiscrete topology showed that topological spaces need not distinguish points. Many familiar theorems from analysis require extra axioms saying that points can be separated by open sets. The first useful condition asks for an open test that sees one point while missing another. [definition: $T_1$ Space] A topological space $(X,\tau)$ is a $T_1$ space if for every pair of distinct points $x,y \in X$, there exist $U,V \in \tau$ such that \begin{align*} x \in U, \qquad y \notin U, \qquad y \in V, \qquad x \notin V. \end{align*} [/definition] The $T_1$ condition prevents the most severe failures of point distinction, but it still does not guarantee disjoint neighbourhoods for distinct points. To recover [uniqueness of limits](/theorems/742), we need a stronger separation condition. [definition: Hausdorff Space] A topological space $(X,\tau)$ is Hausdorff if for every pair of distinct points $x,y \in X$, there exist $U,V \in \tau$ such that \begin{align*} x &\in U, \\ y &\in V, \\ U \cap V &= \varnothing. \end{align*} [/definition] Hausdorffness is designed to make limits behave like they do in metric spaces, but a topological space may have no distance function from which to measure closeness. We therefore need a version of sequential convergence that uses only the open-set data already present in the topology: a sequence should approach $x$ exactly when every open test around $x$ eventually captures all of its terms. [definition: Sequence Convergence] Let $(X,\tau)$ be a topological space, let $(x_n)_{n\in\mathbb N}$ be a sequence in $X$, and let $x \in X$. The sequence $(x_n)_{n\in\mathbb N}$ converges to $x$ if for every neighbourhood $N$ of $x$, there exists $M \in \mathbb N$ such that \begin{align*} x_n \in N \quad \text{for all } n \ge M. \end{align*} [/definition] This definition recovers the usual metric notion when open balls form the neighbourhoods. The theorem below answers the key question raised by the indiscrete example: does disjoint separation of points force a convergent sequence to have only one limit? [quotetheorem:291] This theorem explains why Hausdorffness is built into many definitions of manifolds and geometric spaces. It protects the local-to-global meaning of convergence: a sequence should not be able to settle at two separated destinations at once. ### Countability Axioms Separation tells us whether open sets distinguish points. Countability asks a different question: how much local information is needed to describe all neighbourhoods of a point? [definition: First Countable Space] A topological space $(X,\tau)$ is first countable if for every $x \in X$, there exists a countable family $(U_n)_{n\in\mathbb N}$ of neighbourhoods of $x$ such that for every neighbourhood $N$ of $x$, there exists $n \in \mathbb N$ with \begin{align*} U_n \subset N. \end{align*} [/definition] First countability controls each point separately. For global constructions such as manifolds, we need a countable supply of basic open sets that describes the entire topology at once. [definition: Second Countable Space] A topological space $(X,\tau)$ is second countable if there exists a countable basis $\mathcal B$ for its topology. [/definition] Second countability is a standard hypothesis for manifolds, where it rules out spaces that are locally Euclidean but too large to support the usual analytic tools. Euclidean space supplies the model example. [example: Euclidean Space Is Second Countable] In $\mathbb R^n$ with its usual topology, consider \begin{align*} \mathcal B &= \{B(q,r): q \in \mathbb Q^n,\ r \in \mathbb Q,\ r>0\}. \end{align*} This collection is countable: $\mathbb Q$ is countable, so the finite product $\mathbb Q^n$ is countable, and therefore \begin{align*} \mathbb Q^n\times(\mathbb Q\cap(0,\infty)) \end{align*} is countable. The assignment \begin{align*} (q,r)&\mapsto B(q,r) \end{align*} maps this [countable set](/page/Countable%20Set) onto $\mathcal B$, so $\mathcal B$ is countable. We show that $\mathcal B$ is a basis for the usual topology on $\mathbb R^n$. Each $B(q,r)$ is open in the usual topology, so every union of elements of $\mathcal B$ is usual-open. Conversely, let $U\subset\mathbb R^n$ be usual-open and let $x=(x_1,\ldots,x_n)\in U$. By openness, there is $\varepsilon>0$ such that \begin{align*} B(x,\varepsilon)\subset U. \end{align*} For each coordinate $j$, choose $q_j\in\mathbb Q$ with \begin{align*} |x_j-q_j|<\frac{\varepsilon}{4\sqrt n}. \end{align*} Set $q=(q_1,\ldots,q_n)\in\mathbb Q^n$. Then \begin{align*} \|x-q\|^2 &=\sum_{j=1}^n |x_j-q_j|^2 \\ &<\sum_{j=1}^n \frac{\varepsilon^2}{16n} \\ &=\frac{\varepsilon^2}{16}, \end{align*} so \begin{align*} \|x-q\|<\frac{\varepsilon}{4}. \end{align*} Choose $r\in\mathbb Q$ with \begin{align*} \frac{\varepsilon}{4}<r<\frac{\varepsilon}{2}. \end{align*} Then $x\in B(q,r)$, because \begin{align*} \|x-q\|<\frac{\varepsilon}{4}<r. \end{align*} If $y\in B(q,r)$, then by the triangle inequality, \begin{align*} \|y-x\| &\le \|y-q\|+\|q-x\| \\ &< r+\frac{\varepsilon}{4} \\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{4} \\ &<\varepsilon. \end{align*} Thus $y\in B(x,\varepsilon)$, and hence \begin{align*} B(q,r)\subset B(x,\varepsilon)\subset U. \end{align*} Therefore every point of every usual-open set lies in some member of $\mathcal B$ contained in that open set. Hence every usual-open set is a union of elements of the countable collection $\mathcal B$, so $\mathbb R^n$ is second countable. [/example] Separation and countability often travel together in applications: Hausdorffness makes limits meaningful, while countability makes the topology manageable. ## Compactness and Connectedness ### Compactness Many arguments start with local data, such as open intervals around points or coordinate neighbourhoods on a manifold. Compactness asks when infinitely many local pieces can be replaced by finitely many. [definition: Open Cover] Let $(X,\tau)$ be a topological space and let $A \subset X$. An open cover of $A$ is a family $(U_i)_{i\in I}$ of open subsets of $X$ such that \begin{align*} A &\subset \bigcup_{i\in I} U_i. \end{align*} [/definition] An open cover is only raw local data; by itself it may involve infinitely many sets with no finite control. This becomes a genuine obstruction in analysis: local bounds, local charts, or local extensions may be available around every point, while an argument later needs only finitely many constants or finitely many coordinate patches. Compactness is the condition that turns such pointwise local information into finite global data. [definition: Compact Space] A topological space $(X,\tau)$ is compact if every open cover of $X$ has a finite subcover. [/definition] Compactness generalises closed and bounded intervals without mentioning distance. The next structural question is whether this finite-control property survives continuous maps, since otherwise it would depend on the presentation rather than the topology. [quotetheorem:305] The theorem says compactness is topological rather than metric: continuous maps cannot destroy the finite-subcover property. Its analytic force appears when the target is the real line. [example: Compactness Gives an Extreme Value Theorem] Let $X$ be a nonempty compact topological space and let $f:X\to\mathbb R$ be continuous. Set \begin{align*} A&=f(X). \end{align*} Since $X$ is compact and $f$ is continuous, $A$ is compact in $\mathbb R$ by *Continuous Image of a [Compact Space](/page/Compact%20Space)*. By *Compact Subsets of the Real Line Are Closed and Bounded*, $A$ is closed and bounded. Because $X$ is nonempty, there exists $x_0\in X$, so \begin{align*} f(x_0)\in A, \end{align*} and hence $A$ is nonempty. Since $A$ is nonempty and bounded below, the number \begin{align*} m&=\inf A \end{align*} exists and is finite. For every $\varepsilon>0$, the number $m+\varepsilon$ is not a lower bound for $A$, so there exists $y_\varepsilon\in A$ such that \begin{align*} m\le y_\varepsilon<m+\varepsilon. \end{align*} Also $m-\varepsilon<m\le y_\varepsilon$, so \begin{align*} y_\varepsilon\in A\cap(m-\varepsilon,m+\varepsilon). \end{align*} Thus every open interval around $m$ meets $A$, so $m\in\overline A$. Since $A$ is closed, $\overline A=A$, and therefore \begin{align*} m\in A. \end{align*} By the definition of image, there exists $x_{\min}\in X$ such that \begin{align*} f(x_{\min})&=m=\inf f(X). \end{align*} Similarly, since $A$ is nonempty and bounded above, the number \begin{align*} M&=\sup A \end{align*} exists and is finite. For every $\varepsilon>0$, the number $M-\varepsilon$ is not an upper bound for $A$, so there exists $z_\varepsilon\in A$ such that \begin{align*} M-\varepsilon<z_\varepsilon\le M. \end{align*} Also $z_\varepsilon\le M<M+\varepsilon$, so \begin{align*} z_\varepsilon\in A\cap(M-\varepsilon,M+\varepsilon). \end{align*} Thus every open interval around $M$ meets $A$, so $M\in\overline A=A$. Hence there exists $x_{\max}\in X$ such that \begin{align*} f(x_{\max})&=M=\sup f(X). \end{align*} Therefore a continuous real-valued function on a nonempty compact space attains both its minimum and its maximum. [/example] ### Connectedness Compactness asks whether local covers can be made finite. Connectedness asks whether the space can be separated into two open pieces. This detects whether continuous functions can jump between disconnected regions. [definition: Connected Space] A topological space $(X,\tau)$ is connected if there do not exist nonempty open sets $U,V \in \tau$ such that \begin{align*} X &= U \cup V, \\ U \cap V &= \varnothing. \end{align*} [/definition] Connectedness is qualitative continuity of the space itself. The natural permanence question is whether a continuous image of a connected space can become disconnected. [quotetheorem:296] This theorem is the topological form of the intermediate value principle: continuous maps cannot tear one connected piece into separated pieces. Applying it to maps into $\mathbb R$ gives the familiar conclusion. [example: The Intermediate Value Principle] Let $f:[a,b]\to\mathbb R$ be continuous with the usual topology on $[a,b]$, and let $c$ lie between $f(a)$ and $f(b)$. We show that $c$ is attained by $f$ on $[a,b]$. Set \begin{align*} A&=f([a,b]). \end{align*} The interval $[a,b]$ is connected by *Closed Intervals Are Connected*. Since $f$ is continuous, $A=f([a,b])$ is connected by *Continuous Image of a Connected Space*. By *[Connected Subsets of the Real Line](/theorems/295) Are Intervals*, the connected subset $A\subset\mathbb R$ is an interval. Now \begin{align*} a&\in[a,b],\\ b&\in[a,b], \end{align*} so by the definition of image, \begin{align*} f(a)&\in A,\\ f(b)&\in A. \end{align*} If \begin{align*} f(a)\le c\le f(b), \end{align*} then $c\in A$ because $A$ is an interval containing both endpoints $f(a)$ and $f(b)$. If instead \begin{align*} f(b)\le c\le f(a), \end{align*} the same interval property again gives $c\in A$. Thus in either case $c\in f([a,b])$, so by the definition of image there exists $x\in[a,b]$ such that \begin{align*} f(x)&=c. \end{align*} Thus a continuous real-valued function on an interval attains every value between its endpoint values. [/example] The examples above use compactness and connectedness to force familiar conclusions about real-valued maps, but these properties also serve a classification purpose. To decide whether two spaces have the same topology in disguise, we need properties that survive any reversible continuous change of coordinates. The preceding image theorems give exactly that test when the map has a continuous inverse: a homeomorphism transports compactness and connectedness in both directions, turning them into practical obstructions to spaces being topologically the same. ## Product and Quotient Constructions ### Product Topologies Topology is powerful because it builds new spaces from old ones while prescribing which maps should be continuous. Product spaces solve the problem of making several coordinates vary continuously at once. [definition: Product Topology] Let $((X_i,\tau_i))_{i\in I}$ be a family of topological spaces. The [product topology](/page/Product%20Topology) on \begin{align*} X &= \prod_{i\in I} X_i \end{align*} is the topology generated by the subbasis consisting of all sets \begin{align*} \pi_i^{-1}(U), \end{align*} where $i \in I$, $U \in \tau_i$, and $\pi_i: X \to X_i$ is the coordinate projection. [/definition] The product topology is designed so that every coordinate projection is continuous. To make products usable, we need a test for maps into a product that avoids checking arbitrary unions of finite coordinate constraints directly. [quotetheorem:962] This universal property is the working rule for product spaces: continuity into a product is exactly coordinatewise continuity. It turns an apparently global open-set test into separate tests on the coordinates. ### Quotient Topologies [Quotient topology](/page/Quotient%20Topology) solves the opposite construction problem. Instead of keeping several coordinates, it collapses points according to a surjective map and asks which open sets remain visible after gluing. [definition: Quotient Topology] Let $(X,\tau_X)$ be a topological space, let $Y$ be a set, and let \begin{align*} q: X &\to Y \end{align*} be a surjective map. The quotient topology on $Y$ induced by $q$ is \begin{align*} \tau_Y &= \{V \subset Y : q^{-1}(V) \in \tau_X\}. \end{align*} [/definition] The quotient topology is designed for maps out of a glued space. When the gluing is described by an [equivalence relation](/page/Equivalence%20Relation) $\sim$ on $X$, the quotient set is written $X/{\sim}$ and the quotient projection is the map $\pi:X\to X/{\sim}$ sending each point to its equivalence class. The key question is whether maps out of the quotient can be checked before gluing, where the original topology is already understood. [quotetheorem:1031] This theorem is the formal reason quotient spaces are useful: to define a continuous map out of a glued space, it is enough to define a continuous map before gluing that is constant on the fibres of the quotient map. [example: The Circle as a Quotient Space] Let \begin{align*} q:[0,1]&\to S^1,\\ q(t)&=(\cos(2\pi t),\sin(2\pi t)). \end{align*} We compare the quotient topology $\tau_q$ on $S^1$ induced by $q$ with the usual subspace topology $\tau_u$ inherited from $\mathbb R^2$. First, $q$ is surjective. If $p=(x,y)\in S^1$, then $x^2+y^2=1$, so there is $\theta\in[0,2\pi)$ with \begin{align*} x&=\cos\theta,\\ y&=\sin\theta. \end{align*} Taking $t=\theta/(2\pi)$ gives $t\in[0,1)$ and \begin{align*} q(t)&=(\cos(2\pi t),\sin(2\pi t))\\ &=(\cos\theta,\sin\theta)\\ &=(x,y). \end{align*} The only nontrivial identification is between the endpoints. Indeed, if $q(s)=q(t)$, then \begin{align*} \cos(2\pi s)&=\cos(2\pi t),\\ \sin(2\pi s)&=\sin(2\pi t), \end{align*} so \begin{align*} e^{2\pi i s}&=e^{2\pi i t},\\ e^{2\pi i(s-t)}&=1. \end{align*} Thus $s-t\in\mathbb Z$. Since $s,t\in[0,1]$, we have $s-t\in[-1,1]$, so either $s=t$, or $(s,t)=(1,0)$, or $(s,t)=(0,1)$. Also \begin{align*} q(0)&=(\cos 0,\sin 0)=(1,0),\\ q(1)&=(\cos 2\pi,\sin 2\pi)=(1,0), \end{align*} so $0$ and $1$ are exactly the two distinct points glued together. Now let $V\in\tau_u$. Since the coordinate functions $t\mapsto\cos(2\pi t)$ and $t\mapsto\sin(2\pi t)$ are continuous, $q^{-1}(V)$ is open in $[0,1]$. By the definition of the quotient topology, \begin{align*} V\in\tau_q. \end{align*} Hence $\tau_u\subset\tau_q$. Conversely, let $V\in\tau_q$, so $q^{-1}(V)$ is open in $[0,1]$. We show that $V$ is usual-open in $S^1$. Take $p\in V$. If $p\ne(1,0)$, then $p=q(t_0)$ for a unique $t_0\in(0,1)$. Since $t_0\in q^{-1}(V)$ and $q^{-1}(V)$ is open in $[0,1]$, there is $\delta>0$ such that \begin{align*} (t_0-\delta,t_0+\delta)\subset q^{-1}(V), \end{align*} after decreasing $\delta$ if necessary so that $0<t_0-\delta<t_0+\delta<1$. Set \begin{align*} A&=q((t_0-\delta,t_0+\delta)). \end{align*} Then $p\in A\subset V$. This set is an open arc in the usual subspace topology on $S^1$: around each of its points $q(t)$ with $t_0-\delta<t<t_0+\delta$, choose $\eta>0$ with \begin{align*} (t-\eta,t+\eta)\subset(t_0-\delta,t_0+\delta), \end{align*} and the image $q((t-\eta,t+\eta))$ is the corresponding small usual open arc in $S^1$. If $p=(1,0)$, then $0,1\in q^{-1}(V)$. Since $q^{-1}(V)$ is open in $[0,1]$, there is $\varepsilon>0$ such that \begin{align*} [0,\varepsilon)\cup(1-\varepsilon,1]\subset q^{-1}(V). \end{align*} Therefore \begin{align*} q([0,\varepsilon)\cup(1-\varepsilon,1])\subset V. \end{align*} This image is the usual open arc around $(1,0)$ crossing the glued endpoint, because it consists exactly of points \begin{align*} (\cos(2\pi t),\sin(2\pi t)) \end{align*} with $t$ close to $0$ or close to $1$, which are precisely the points of $S^1$ with angle close to $0$ modulo $2\pi$. Every point of $V$ therefore has a usual-open arc contained in $V$, so $V\in\tau_u$. Hence $\tau_q\subset\tau_u$, and therefore \begin{align*} \tau_q&=\tau_u. \end{align*} Thus the quotient topology on $S^1$ induced by $q$ is exactly the usual topology on the unit circle: the interval has been turned into a circle by identifying only $0$ with $1$, and the open arcs crossing $(1,0)$ are recorded by neighbourhoods of both endpoints in $[0,1]$. [/example] Products and quotients are complementary: products assemble independent coordinates, while quotients impose identifications. Much of topology consists of understanding which properties survive these constructions. ## Beyond and Connected Topics Topological spaces are the entry point to [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), where metric spaces, compactness, connectedness, and continuity are developed into the standard toolkit for analysis. That course-level perspective explains how the abstract definitions recover the theorems of real analysis. Algebraic topology begins when topological spaces are studied through invariants such as homotopy groups, homology groups, and cohomology rings. [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology) develops the first systematic invariants, while [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology) continues toward deeper structural tools. Algebraic geometry uses topology in a different direction. The Zariski topology on affine varieties and schemes is much coarser than the Euclidean topology, but it is tuned to polynomial equations. This makes [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry) a natural continuation for seeing how topological language adapts to algebraic objects. Functional analysis and PDE use topologies on function spaces to encode convergence weaker than norm convergence. Weak topologies, compact embeddings, and distribution spaces show that choosing the right topology is often the decisive step in an existence theorem. Differential geometry adds smooth structure to topology. A manifold is first a topological space locally modelled on Euclidean space, and then an atlas specifies which coordinate changes are smooth. The topological assumptions ensure that local coordinate data glue into a usable global object. ## References Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). Androma, [Cambridge II Algebraic Geometry](/page/Cambridge%20II%20Algebraic%20Geometry). James R. Munkres, *Topology* (2000). John L. Kelley, *General Topology* (1955). Allen Hatcher, *Algebraic Topology* (2002).