## Closed Sets & Limit Points

July 2020

Problem 5: Let $$X$$ be an ordered set in the order topology. Show that $$\overline{(a, b)} \subset [a, b]$$. Under what conditions does the equality hold?

Theorem 17.6 in the book states that, given a subset $$A$$ of $$X$$, and given the set $$A'$$ of all limit points of $$A$$, $$\overline{A} = A \cup A'$$. $$(a, b)$$ is clearly in $$[a, b]$$, so we just need to prove that if $$x$$ is a limit point of $$(a,b)$$, it is in $$[a, b]$$.

Let $$x$$ be a limit point of $$(a, b)$$, and suppose it is not in $$[a, b]$$. Then, either $$x < a$$ or $$x > b$$. This would allow for the existence of neighbourhood(s) of $$x$$ (such as $$\{p | p< a\}$$ if $$x < a$$ and $$\{p | p>b\}$$ if $$x>b$$) that do not intersect any points of
$$(a, b)$$, which contradicts our assumption that $$x$$ is a limit point. Thus, any limit point of $$(a, b)$$ must be in $$[a, b]$$. $$\tag*{\blacksquare}$$

Problem 12: Show that a subspace of a Hausdorff space is Hausdorff.

Given a Hausdorff space $$X$$, and $$A \subset X$$, the subspace topology on $$A$$ is defined as the collection of all sets $$A \cap Y$$ where $$Y$$ is open in $$X$$. Given $$x, y \in A \subset X$$, since $$X$$ is a Hausdorff space we can define a neighbourhood $$U_1$$ of $$x$$ in $$X$$ and a neighbourhood $$U_2$$ of $$y$$ in $$X$$ such that the two neighbourhoods are disjoint. If $$U_1 \cap U_2 = \varnothing$$, then $$U_1 \cap A \cap U_2 = \varnothing$$, and so there are disjoint neighbourhoods of $$x$$ and $$y$$ in the subspace topology on $$A$$, making it a Hausdorff space. $$\tag*{\blacksquare}$$