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]