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?