# How to prove that $f:[0,1]\cup[2,4]\rightarrow \mathbb R$ is continuous using topology?

I am trying to understand the following:

Theorem: Let $$(X,d_X)$$ and $$Y(d_Y)$$ be metric spaces. A function $$f:X\rightarrow Y$$ is continuous if and only if for every open $$U\subset Y$$, then $$f^{-1}(U) = \{x\in X : f(x)\in U\}$$ is open in $$X$$.

If I want to show that the function $$f:[0,1]\cup[2,4]\subset \mathbb R \rightarrow \mathbb R$$ defined by $$f(x) =\begin{cases} 1, & x\in [0,1];\\ 2, & x\in [2,4]. \end{cases}$$ is continuous, then I need to consider any open set $$U\subset \mathbb R$$ and show that $$f^{-1}(U) = \{x\in [0,1]\cup[2,4] : f(x)\in U\}$$ is open.

I understand that if $$1 \notin U$$ and $$2 \notin U$$, then $$f^{-1}(U) = \emptyset$$. But I don't understand what happens when only $$1\in U$$, or only $$2\in U$$, or $$1,2\in U$$.

If only $$1\in U$$, then is it correct to say that $$f^{-1}(U)=[0,1]$$? If only $$2\in U$$, then is it correct to say that $$f^{-1}(U)=[2,4]$$? But I am really confused since these are closed intervals. I am lost.

Can someone help me to find $$f^{-1}(U)$$ for those cases? Any clues or hints will be appreciated.

• Everything you state is correct. But the fact that $[0,1]$ and $[2,4]$ are closed intervals in $\mathbb{R}$ does not mean they are not open sets in $[0,1] \cup [2,4]$. Are you familiar with the definition of subspace topology? Mar 20, 2022 at 9:44
• $[0,1]$ and $[2,4]$ are open in $[0,1] \cup [2,4]$, since we consider this space to carry the subspace topology inherited from $\mathbb{R}$. For instance, $[0,1] = [0,1] \cap \mathbb{R}$, so it is open. Mar 20, 2022 at 9:45
You are missing the fact that $$[1,2]$$ and $$[2,4]$$ are actually open sets in $$[0,1]\cup [2,4]$$. For example, $$[0,1]=\{x\in [0,1]\cup [2,4]: |x-0| <1.5\}$$.
So $$1 \in U , 2 \notin U$$ gives $$f^{-1}(U)=[0,1]$$ which is open.