[proofplan] We prove convergence of the image sequence directly from the neighbourhood definition of convergence in a [topological space](/page/Topological%20Space). Given an arbitrary open neighbourhood of $f(x)$ in $Y$, continuity of $f$ pulls it back to an open neighbourhood of $x$ in $X$. Since $x_k\to x$, the sequence is eventually in this preimage, and therefore its image is eventually in the original neighbourhood. [/proofplan] [step:Pull back an arbitrary neighbourhood of $f(x)$] Let $V\in\tau_Y$ be an [open set](/page/Open%20Set) such that $f(x)\in V$. Since $f:X\to Y$ is continuous, the preimage \begin{align*} f^{-1}(V):=\{z\in X: f(z)\in V\} \end{align*} belongs to $\tau_X$. Since $f(x)\in V$, we have $x\in f^{-1}(V)$, so $f^{-1}(V)$ is an open neighbourhood of $x$ in $(X,\tau_X)$. [guided] To prove $f(x_k)\to f(x)$ in $Y$, we must start with an arbitrary open neighbourhood of the proposed limit $f(x)$. Let $V\in\tau_Y$ be open and suppose $f(x)\in V$. Continuity is exactly the property that open sets in the codomain pull back to open sets in the domain. Therefore the set \begin{align*} f^{-1}(V):=\{z\in X: f(z)\in V\} \end{align*} is an element of $\tau_X$. This set contains $x$, because $f(x)\in V$. Hence $f^{-1}(V)$ is an open neighbourhood of $x$ in the topological space $(X,\tau_X)$. [/guided] [/step] [step:Use convergence of $x_k$ to force eventual membership in the preimage] Since $x_k\to x$ in $(X,\tau_X)$ and $f^{-1}(V)$ is an open neighbourhood of $x$, there exists $N\in\mathbb{N}$ such that for every $k\in\mathbb{N}$ with $k\ge N$, \begin{align*} x_k\in f^{-1}(V). \end{align*} By the definition of preimage, this implies that for every $k\ge N$, \begin{align*} f(x_k)\in V. \end{align*} [guided] Now we use the hypothesis $x_k\to x$. In a topological space, this means that for every open neighbourhood $U$ of $x$, there exists $N\in\mathbb{N}$ such that $x_k\in U$ for every $k\ge N$. We apply this definition to the particular open neighbourhood \begin{align*} U:=f^{-1}(V). \end{align*} The previous step verified that $U\in\tau_X$ and $x\in U$, so the convergence hypothesis gives an index $N\in\mathbb{N}$ such that \begin{align*} x_k\in f^{-1}(V) \end{align*} for every $k\in\mathbb{N}$ with $k\ge N$. Unwinding the definition of preimage, $x_k\in f^{-1}(V)$ is equivalent to $f(x_k)\in V$. Therefore \begin{align*} f(x_k)\in V \end{align*} for every $k\ge N$. [/guided] [/step] [step:Conclude convergence of the image sequence] We have shown that for every open set $V\in\tau_Y$ with $f(x)\in V$, there exists $N\in\mathbb{N}$ such that $f(x_k)\in V$ for every $k\ge N$. By the definition of convergence of a sequence in the topological space $(Y,\tau_Y)$, this proves \begin{align*} f(x_k)\to f(x) \end{align*} in $Y$. [/step]