# Show that $f:= id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R}, \tau_d)$ is not continuous

I've got to show that the function $$f:= id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R}, \tau_d)$$ is not continuous.

$$(\mathbb{R}, \tau), (\mathbb{R}, \tau_d)$$ are both topologies. $$\tau = \{\emptyset\} \cup \{U \subseteq \mathbb{R}: \mathbb{R}\setminus U \text{ is countable}\}$$ and $$\tau_d$$ is the discrete topology.

$$id$$ is the identity-map.

Maybe I wrongfully understand the problem but it seems really easy to me.

For $$f$$ to be continuous, I need to show that $$f^{-1}(V)$$ is open in $$\tau$$ for any $$V\subseteq \mathbb{R}$$.

Let $$x\in \mathbb{R}$$. Since $$\tau_d$$ is the discrete topology, $$\{x\} \subseteq \mathbb{R}$$ is open in $$\tau_d$$. But $$f^{-1}(\{x\})=\{x\} \notin \tau$$ since $$\mathbb{R}\setminus \{x\}$$ is not countable.

Therefore $$f$$ cannot be continuous.

Am I correct?

Note as an immediate corollary, that any function is continuous if we endow the domain $$\mathbb{R}$$ of the function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ with the discrete topology.