Is $I:\mathbb{R}\rightarrow\mathbb{Z},x\mapsto \lfloor x \rfloor$ continuous? $\mathbb{Z}$ is endowed with the subspace topology.

Question

I'd like to say that it isn't.

Consider the set $\{1\}$ in $\mathbb{Z}$, where $k$ is some integer. It is open because it equals $(0.5,1.5)\cap \mathbb{Z}$, and $(0.5,1.5)$ is open in $\mathbb{R}$.

If we take $I^{-1}(\{1\})$, we get $[1,2)$, which isn't open in $\mathbb{R}$. Hence, the function isn't continuous.

Answer 1

You're right.

More generally, let $X$ be connected and $Y$ be discrete. A map $f: X \rightarrow Y$ is continuous if and only if it is constant.