**
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 \([a, b]\).

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 \([a, b]\).

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.