Why is a metric space an open subset of itself?

Question

I've been reading about topology, and I've come across the following:

Given a metric space $X$, the entire space $X$ is an open subset of $X$.

I'm having some trouble thinking about this. I have a counterexample in mind. Suppose we consider the closed subset of $\mathbb{R}$ called $A = [a, b] \in \mathbb{R}$. Isn't $A$ also a metric space? As far as I know, the answer is yes. So, based on the above, $A$ is an open subset of $A$. But, since $a$ and $b$ aren't interior points of $A$, how is $A$ an open subset?

What is wrong with my line of thinking?

Answer 1

What you’re missing is that in the space $A$ the points $a$ and $b$ are interior points of $A$, even though they are not interior points of $A$ in the space $\Bbb R$. Let $r=b-a$. By definition the open ball of radius $r$ centred at $a$ in the space $A$ is $\{x\in A:|x-a|<r\}$, the set of points in $A$ that are less than distance $r$ from $a$. And

$$\{x\in A:|x-a|<r\}=[a,b)\subseteq A\;,$$

so $a$ is in the interior of $A$ in the space $A$: the open ball of radius $r$ around $a$ is contained in $A$.

You can do the same thing at $b$.

Answer 2

By definition, $A$ is an open (and also a closed) subset of the metric space $A$ (endowed with a topology). This is one of the axioms defining a topology.

Now, if you look at a small open ball (in $A$) centered on $a$, it will be included in $A$. The reason is that such open balls will be of the form $(a-\epsilon,a+\epsilon)\cap A = [a,a+\epsilon)$. So $a$ is an interior point: there is no inconsistency between the general definition of the topology and the specific (equivalent) definition for metric spaces.

Answer 3

Given a non-empty set $A$ endowed with a metric $d$, we define the metric space as $X:=(A,d)$.

For each $p\in A$, the open ball centred at point $p$ with radius $r$ is given by $N_p(r):=\{ q\in A:d(p,q)<r,r\in\mathbb{R}_{>0} \}\subset A$, implying that $A$ is open relative to itself.