Why is $[0,1]$ an open subset of $[0,1] \cup [2,3]$?

Question

Given a metric space $X = [0,1]\cup[2,3]$

I have to show $[0,1]$ is both open and closed in $X$.

This question is also asked in this thread :

Let $X = [0,1] \cup [2,3]$ be a metric space. Why is $[0,1]$ both open and closed?

I understand why $[0,1]$ is closed, but am having trouble understanding why it is open.

From the answer in the previous thread, I understand that $[0,1]$ is open as it is the complement of $[2,3]$ which is closed.

But for the point $0$ in $[0,1]$ we can construct no open ball which lies in X. So then how is $[0,1]$ open in X?

Answer 1

Thanks a lot MichaelBurr and copper.hat.

The key was to think in terms of the relative metric for $X$ and not the metric used for $\mathbb{R}$. The relative metric restricts the definition of the metric for only those points which belong to $[0,1]\cup[2,3]$. In this case, $[0,1/2)$ is also an open ball in [0,1] when X is the metric space under consideration.

Alternatively,as suggested by copper.hat, one may look at $[0,1]$ as $X\cap[0,1]$ where $[0,1]$ is closed in $\mathbb{R}$. Also it can be seen as $X\cap(-1,3/2)$ where $(-1,3/2)$ is open in $\mathbb{R}$.

Answer 2

$[0,\frac{1}{2})$ is an open ball in $X$ (not in $\mathbb R$):

$$[0,\frac{1}{2}) = \{ x \in X : |x| < \frac{1}{2} \}$$