Closed Sets and Limit Points

Adar Kahiri
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 \([a, b]\).

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.