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?
Best Answer
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 $b$.