Let $(X,d)$ be a metric space and $Y \subseteq X$ a subset. I want to show the following without using accumulation points or limit points at all.
- $\overline Y$ is a closed subset of $X$.
- $\overline Y$ is contained in every closed set which contains $Y$.
My definitions are:
The interior of $Y$ is
$$Y^\circ = \{y \in Y \mid \exists \varepsilon \gt 0:B_\varepsilon(y) \subseteq Y\}.$$
The boundary of $Y$ is
$$ \partial Y= \{x \in X \mid \forall \varepsilon \gt 0: B_\varepsilon(x) \cap Y \neq \emptyset \neq B_\varepsilon(x) \cap (X \setminus Y) \}.$$
And the closure of $Y$ is
$$\overline Y = Y \cup \partial Y.$$
I know also that a set $O \subseteq X$ is open in $X$ iff every convergent sequence with limit in $O$ has also almost all of its terms in $O$. And a set $A$ is closed in $X$ iff every convergent sequence with all of its terms in $A$ has also its limit in $A$.
I tried showing that $X \setminus \overline Y$ is open as well as using the above result for sequences but both times I got stuck in complicated set expressions resulting from figuring out what $X \setminus \overline Y$ is. Can you show me how to prove this?
I found only proofs using limit or accumulation points (links: here, here, here, here, here)
Best Answer