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

In learning about connected metric spaces, I came across the definition that a metric space $X$ is not connected if there exists an open and closed subset of $X$.

In the above example, I can think of two reasons why $[0,1]$ is both open and closed but I don't know which is the corrected interpretation.

1) $[0,1]$ is an open subset with respect to $\mathbb{R}$ but a closed subset with respect to $X$. This shows it is both open and closed.

2) $[0,1]$ is closed in $X$ $\implies$ $[2,3]$ is open. But we also have that $[2,3]$ is closed in $X$ $\implies$ $[0,1]$ is open. Hence, $[0,1]$ is both open and closed.

Are either of these the correct interpretation? Which one is wrong and why?

• You're not telling us the topology $X$ has. Presumably it is the subspace topology inherited from $\Bbb R$. – Pedro Tamaroff Nov 2 '14 at 20:17

(2) is correct, but (1) is not -- the set $[0,1]$ is certainly not open in $\mathbb R$.

Both of these intervals are open in $X$ because they each are the union of open balls (in $X$!) of radius $1$ about each of their points. Both are also closed in $X$, each being the complement of an open set in $X$.

If you use the induced topology from $\mathbb{R}$, which is likely what you do, then a subset $U$ of $X$ is open in $X$ if there is an open subset $U'$ of $\mathbb{R}$ such that $U' \cap X = U$.

Now try to find an open subset of $\mathbb{R}$ so that the intersection of $X$ with it is $[0,1]$. For example, take a slightly larger open interval.

In this way you show $[0,1]$ is open in $X$ (it is in fact not open in $\mathbb{R}$). Then continue as you did.

For one thing, $X$ is not connected. Depending on what definition you are using, $X$ is disconnected if it is the union of two disjoint clopen (open and closed) sets.

Clearly, $[0,1] \cap[2,3] = \emptyset$, so both $[0,1]$ and $[2,3]$ are clopen under the usual topology.

Without any loss of generality, taking $[0,1]$ we get $$[0,1] = X \cap [0,1]$$

where $[0,1]$ is closed in $\Bbb R$, so $[0,1]$ is closed in $X$.

We also have, $$[0,1] = X \cap (-1, 1.5)$$

where $(-1,1.5)$ is open in $\Bbb R$. So $[0,1]$ is open in $X$ as well.