[proofplan] We prove continuity by using the open-set definition. An arbitrary [open set](/page/Open%20Set) $U \in \tau_Y$ has preimage $f^{-1}(U) \subset X$. Since $\tau_X$ is the [discrete topology](/page/Discrete%20Topology), every subset of $X$ is open, so $f^{-1}(U) \in \tau_X$. This verifies the defining condition for continuity. [/proofplan] [step:Check that every open set in the target has open preimage] Let $U \in \tau_Y$ be arbitrary. The preimage of $U$ under $f$ is the subset of $X$ defined by \begin{align*} f^{-1}(U) = \{x \in X : f(x) \in U\}. \end{align*} Since $\tau_X = \mathcal{P}(X)$, every subset of $X$ belongs to $\tau_X$. Hence $f^{-1}(U) \in \tau_X$. [guided] To prove that $f: (X, \tau_X) \to (Y, \tau_Y)$ is continuous, we use the definition of continuity between topological spaces: for every open set $U \in \tau_Y$, the preimage $f^{-1}(U)$ must be open in $(X, \tau_X)$. Let $U \in \tau_Y$ be arbitrary. The preimage of $U$ under the function $f: X \to Y$ is \begin{align*} f^{-1}(U) = \{x \in X : f(x) \in U\}. \end{align*} This is a subset of $X$. The defining property of the discrete topology is that every subset of $X$ is open; equivalently, $\tau_X = \mathcal{P}(X)$. Therefore $f^{-1}(U) \in \tau_X$. Because the choice of $U \in \tau_Y$ was arbitrary, every open subset of the target has open preimage in the domain. [/guided] [/step] [step:Conclude continuity from the open-set criterion] The preceding step proves that for every $U \in \tau_Y$, one has $f^{-1}(U) \in \tau_X$. By the definition of continuity for maps between topological spaces, $f: (X, \tau_X) \to (Y, \tau_Y)$ is continuous. [/step]