[proofplan] We prove continuity by using the open-set definition: a map between topological spaces is continuous exactly when the preimage of every open set in the codomain is open in the domain. Since the topology on $X$ is discrete, every subset of $X$ is open. Therefore each preimage $f^{-1}(V)\subseteq X$ of an open set $V\subseteq Y$ automatically belongs to the topology on $X$. [/proofplan] [step:Declare the topology on the domain as the full power set] Let $\tau_X$ denote the topology on $X$. Since $X$ has the [discrete topology](/page/Discrete%20Topology), by definition \begin{align*} \tau_X = \mathcal{P}(X), \end{align*} where $\mathcal{P}(X)$ denotes the power set of $X$. Hence every subset $A\subseteq X$ is an open set in $(X,\tau_X)$. [/step] [step:Show that preimages of open sets in $Y$ are open in $X$] Let $V\in \tau_Y$ be an arbitrary open set of the topological space $(Y,\tau_Y)$. Define the preimage subset \begin{align*} f^{-1}(V) := \{x\in X : f(x)\in V\}. \end{align*} This is a subset of $X$, so $f^{-1}(V)\in \mathcal{P}(X)=\tau_X$. Thus $f^{-1}(V)$ is open in $(X,\tau_X)$. [guided] We want to prove that the map \begin{align*} f: (X,\tau_X) &\to (Y,\tau_Y) \end{align*} is continuous. By the open-set definition of [continuity](/page/Continuity), it is enough to prove the following statement: for every open set $V\in \tau_Y$, the preimage $f^{-1}(V)$ belongs to $\tau_X$. So let $V\in \tau_Y$ be arbitrary. The preimage of $V$ under $f$ is the subset of $X$ defined by \begin{align*} f^{-1}(V) := \{x\in X : f(x)\in V\}. \end{align*} The important point is that this set is a subset of the domain $X$. Since $X$ carries the discrete topology, its topology is \begin{align*} \tau_X = \mathcal{P}(X). \end{align*} Thus every subset of $X$ is open in $(X,\tau_X)$. Applying this to the particular subset $f^{-1}(V)\subseteq X$, we get \begin{align*} f^{-1}(V)\in \tau_X. \end{align*} Since $V\in \tau_Y$ was arbitrary, the preimage of every open set in $Y$ is open in $X$. [/guided] [/step] [step:Conclude continuity from the open-set criterion] The preceding step proves that for every $V\in \tau_Y$, the preimage $f^{-1}(V)$ belongs to $\tau_X$. By the open-set criterion for [continuity](/page/Continuity), the map \begin{align*} f: (X,\tau_X) &\to (Y,\tau_Y) \end{align*} is continuous. This proves the theorem. [/step]