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?


You are indeed correct. This is a typical example to show how common functions that in highschool we thought to be obviously continuous, may in fact be not continuous if we change the topologies of the domain and the codomain. It is to show how how continuity is a property that crucially depends on the open sets, besides the function as well.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.