Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. Let $A \subset Y$, and equip $A$ with the [subspace topology](/page/Subspace%20Topology) \begin{align*} \tau_A := \{A \cap V : V \in \tau_Y\}. \end{align*} Let $f: X \to A$ be a function, and let $i: A \to Y$ be the inclusion map, defined by $i(a)=a$ for every $a \in A$. Then $f: (X,\tau_X) \to (A,\tau_A)$ is continuous if and only if $i \circ f: (X,\tau_X) \to (Y,\tau_Y)$ is continuous.

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces, let $f:X\to Y$ be continuous, and let $(x_k)_{k\in\mathbb{N}}$ be a sequence in $X$. If $x\in X$ and $x_k\to x$ in $(X,\tau_X)$, then $f(x_k)\to f(x)$ in $(Y,\tau_Y)$.

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $\tau_X$ and $\tau_Y$ be the metric topologies induced by $d_X$ and $d_Y$, respectively. Let $f: X \to Y$ be a map, and let $x_0 \in X$. Then $f$ is continuous at $x_0$ with respect to the topologies $\tau_X$ and $\tau_Y$ if and only if for every $\varepsilon > 0$ there exists $\delta > 0$ such that, for every $x \in X$, \begin{align*} d_X(x,x_0)<\delta \implies d_Y(f(x),f(x_0))<\varepsilon. \end{align*}

Let $(X,\tau_X)$ be a topological space, let $Y$ be a set equipped with the indiscrete topology $\tau_Y=\{\varnothing,Y\}$, and let $f:X\to Y$ be a function. Then $f:(X,\tau_X)\to (Y,\tau_Y)$ is continuous.

Let $(X,\tau_X)$ be a first-countable [topological space](/page/Topological%20Space), let $(Y,\tau_Y)$ be a topological space, and let $f:X\to Y$ be a function. Then $f$ is continuous if and only if, for every point $x\in X$ and every sequence $(x_k)_{k\in\mathbb N}$ in $X$, convergence $x_k\to x$ in $(X,\tau_X)$ implies convergence $f(x_k)\to f(x)$ in $(Y,\tau_Y)$.

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces, and let $f: X \to Y$ be a function. Then $f$ is continuous, meaning that $f^{-1}(V) \in \tau_X$ for every $V \in \tau_Y$, if and only if $f$ is continuous at every point $x_0 \in X$, meaning that for every $V \in \tau_Y$ with $f(x_0) \in V$, there exists $U \in \tau_X$ such that $x_0 \in U$ and $f(U) \subset V$.

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $A\in\mathcal F$ be an event. If $A^c:=\Omega\setminus A$ denotes the complement of $A$ in $\Omega$, then \begin{align*} \mathbb P(A^c)=1-\mathbb P(A). \end{align*}

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $A,B\in\mathcal F$ be events. Then \begin{align*} \mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B). \end{align*}

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, and let $B \in \mathcal F$ satisfy $\mathbb P(B)>0$. Define the map \begin{align*} \mathbb P_B:\mathcal F &\to [0,1] \\ A &\mapsto \mathbb P(A \mid B) := \frac{\mathbb P(A \cap B)}{\mathbb P(B)}. \end{align*} Then $\mathbb P_B$ is a probability measure on $(\Omega,\mathcal F)$. Equivalently, $(\Omega,\mathcal F,\mathbb P_B)$ is a probability space.

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. Let $A,B\in\mathcal F$ be events such that $A\cap B=\varnothing$, $\mathbb P(A)>0$, and $\mathbb P(B)>0$. Then $A$ and $B$ are not independent, meaning that $\mathbb P(A\cap B)\neq \mathbb P(A)\mathbb P(B)$.