Showing that $[0,1]$ is open in $[0,1]\cup[2,3]$ (euclidean distance)

Question

Let $M:=[0,1] \cup [2,3]$ be a metric space.

How to show that $[0,1]$ is open and closed under the euclidean metric $||x-y||=\sqrt{(x-y)^2}$?

My idea was:

To show that $[0,1]$ is open in $M$, I have to show that

$\forall x \in [0,1] \exists \epsilon >0: \lbrace y \in [0,1]\cup [2,3]: ||x-y||<\epsilon \rbrace \subset [0,1]$

But if I take $x=0$ and $\epsilon =4$, I get $||0-3||=3<4$, so $\lbrace3\rbrace \not \subset [0,1] $

How can it be shown correctly?

You have to show that there is some $\varepsilon$ which does the job. Why are you choosing $4$? — Xander Henderson, Oct 04 '22 at 16:05
Consider the $\epsilon$ "ball" around some $x\in[0,1]$. It's just $(x-\epsilon,x+\epsilon)$ normally, but that is not the case here since we are restricted to the parent topological space $[0,1]\cup[2,3]$ so the actual $\epsilon$ ball around $x$ is $(x-\epsilon,x+\epsilon)\cap([0,1]\cup[2,3])$. Consider the case: $0<\epsilon<1$. — jdods, Oct 04 '22 at 16:13

score 1 · Accepted Answer · answered Oct 04 '22 at 16:18

For each $x\in[0,1]$, take $\varepsilon=1$. Than $(x-1,x+1)\cap[0,1]\subset[0,1]$. This proves that $[0,1]$ is open in $[0,1]\cup[2,3]$.

And it is closed in$[0,1]\cup[2,3]$ since its complement in $[0,1]\cup[2,3]$ is $[2,3]$, which is open, for the same reason.

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