I am studying topology, on my own, using a text I found online. I am currently reviewing the “Metrics” section that reminds me of the real analysis course I took over 10 years ago.
The text ask me to “show” the following:
Suppose M is a metric space. Show that an open ball in M is an open subset, and a closed ball in M is a closed subset.
I have what I think is a counterexample to the second part. First, let me state the definitions as they are written in the book I am using:
For any $x \in M$ and $r>0$, the (open) ball of radius r around x is the set
$$ B_r(x)=\{y \in M: d(x,y)<r \}, $$
and the closed ball of radius r around x is
$$ \overline B_r(x)=\{y \in M: d(x,y) \leq r \}, $$A subset $A \subseteq M $ is said to be an open subset of M if it contains an open ball around each of its points.
A subset $A \subseteq M $ is said to be an closed subset of M if M\A is open.
I believe the following is a counterexample to this:
Let $$M = [1,10].$$ Now $ \overline B_1(5)=\{y \in M: d(5,y) \leq 1 \} $ is a closed ball. More simply put, $ \overline B_1(5)=[4,6] $. Lets call the closed ball $A$.
$$ A=\overline B_1(5)=[4,6]$$
Clearly, $A \subseteq M $, and $ M-A = [1,4) \cup (6,10] $. However $M-A$ is not open because $\{1\}$ and $\{10\}$ cannot have open balls around them without going beyond M.
Is there an error in the text, or an error in my thinking?
Best Answer
Your intuition is good in that $[1,4)$ and $(6,10]$ are not open in $\mathbb R$. However, they are open in $M=[1,10]$ - that is, openness is a property relative to the metric space. In fact, consider the definition of an open ball of radius $3$ around $1$ in $M$: $$B_3(1)=\{y\in M:|y-1|<3\}$$ the condition $y\in M$ means that $y$ must be in $[1,10]$ and the condition $|y-1|<3$ implies that $y$ cannot be $4$ or more - thus $B_3(1)=[1,4)$.
Essentially, when we're working in a subspace of a metric space, we can "chop off" open sets like this. That is, we have that $(-2,4)$ was an open ball, and when we're in the metric space on $[1,10]$ we take their intersection to get $[1,4)$ being open in the subspace. We do not have to worry about the exclusion of points which are not in the space.