Example of an identity function that's not continuous I was looking at this big list mathoverflow question about common misconceptions: https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics specifically as a comment to this question
something brought up in the comments but not elaborated on was that the identity function is not necessarily continuous. I haven't been able to find anything about it, and I can't think of any examples for this. If anyone could offer some insight on this, I am interested to hear.
 A: Consider two different topologies $ \tau_{1} $, $ \tau_{2} $ on the same set $X$. It can be easily shown (just apply the definition of a continuous map) that the identity map $id_{X}:(X, \tau_{2}) \rightarrow (X, \tau_{1})$ is a continuous function if and only if $\tau_{1} \subseteq \tau_{2}$. 
P.S.: Note that in the above, the term "identity" map for $id_{X}:id_{X}(x)=x, \ \forall x \in X$ is in fact used only in the set-theoretic sense but not in the topological sense.
A: Consider $X=(\mathbb{R}, \textrm{indiscrete topology})$ and $Y=(\mathbb{R}, \textrm{euclidean topology})$. Let $\iota: X \to Y$ be the map given by $\iota(x) = x$. Observe that $\iota^{-1}(\{z\}) = \{z\}$ and $\{z\}$ is closed in $Y$, but not in $X$ since the only closed sets are $\emptyset, \mathbb{R}$ with the indiscrete topology. 
A: Just to give another example in addition to Faraad, you can take the identity from from $\mathbb{R}$ with the usual topology into $\mathbb{R}$ with the discrete topology. These two types of examples are very useful. Every function whose domain has the discrete topology is continuous. in the same way it is "hard" to have a continuous function into the discrete topology. Similarly every function into the trivial (indiscrete) topology is continuous, it is "hard" to have a continuous function from the trivial topology.
A: I would like to contribute to this discussion with another example.
Let $X_1=C^0([0,1])$ with the norm $\|f\|_1=\int_0^1|f(x)|\,dx$ and let $X_\infty=C^0([0,1])$ with the norm $\|f\|_\infty=\sup_{x\in[0,1]}|f(x)|$. It can be proven that both are normed spaces. They are the same set but endowed with different norms, that is, different topologies.
Let $I_d:X_1\longrightarrow X_\infty$ be the identity mapping, that is, $I_d(f)=f$. Since $I_d$ is a linear operator, we know that continuity is equivalent to boundedness (see here).
Suppose $I_d$ is bounded. Then there exists $c>0$ such that
$$\|f\|_\infty=\|I_d(f)\|_\infty\leq c\|f\|_1,\ \forall f\in C^0([0,1]).$$
However, let us consider $f_n:[0,1]\longrightarrow \mathbb{R}$ the following continuous function

We can see that
$$n=\|f_n\|_\infty\leq c\|f_n\|_1=\frac{c}{2},\ \forall n\in \mathbb{N},$$
which is an absurd!
