#2. Show that if $A$ is closed in $Y$ and $Y$ is closed in $X$, then $A$ is closed in $X$.

Since $A$ is closed in $Y$, we have $A = Y - B$ for some open $B$ in $Y$. Since $Y$ has the subspace topology, $B = Y \cap U$ for some open $U$ in $X$, so

$A = Y - B = Y - (U \cap Y) = Y - U$.

But since $Y$ is closed in $X$, we have $Y = X - W$ for some $W$ open in $X$. Thus

$A = Y - U = (X - W) - U = X - (W \cup U)$

where $W$ and $U$ are open in $X$. Thus $A$ is the complement of an open set in $X$, so $A$ is closed in $X$.

Alternative proof:

By theorem 17.2, $A$ is equal to the intersection of some closed set $C$ of $X$ with $Y$. Then $A = C \cap Y$ where both $Y$ and $C$ are closed in $X$, and the intersection of closed sets is closed, so $A$ is closed in $X$.

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

We denote the set of all limit points of $(a, b)$ with $(a, b)'$. Then $\overline{(a, b)} = (a, b) \cup (a, b)'$ per theorem 17.6. Clearly $(a, b) \subset [a, b]$, so we just need to show that $(a, b)' \subset [a, b]$.

Suppose $x \in (a, b)'$. If $x < a$, then the open ray $(-\infty, a)$ is a neighborhood of $x$ that is disjoint from $(a, b)$, meaning $x$ cannot be a limit point of $(a, b)$. Likewise if $x > b$, then the open ray $(b, +\infty)$ is a neighborhood of $x$ disjoint from $(a, b)$, so $x$ again cannot be a limit point of $(a, b)$. Thus if $x \in (a, b)'$, then $x \nless a$ and $x \ngtr b$, so $x \in [a, b]$.

Thus $(a, b)' \subset [a, b]$, so $\overline{(a, b)} \subset [a, b]$.

Equality holds if and only if $a$ and $b$ are limit points of $(a, b)$. Note that $a$ and $b$ are limit points of $(a, b)$ if and only if $b$ does not have an immediate predecessor and $a$ does not have an immediate successor.

For if $b$ has an immediate predecessor $j$, then the open ray $(j, +\infty)$ is a neighborhood of $b$ that is disjoint from $(a, b)$, so $b$ is not a limit point of $(a, b)$. A similar argument applies if $a$ has an immediate successor.

Conversely, if $b$ has no immediate predecessor, then consider any neighborhood $U$ of $b$. $U$ must contain a basis element (i.e. an open interval) $B$ around $b$, its left end defined by some $j < b$. Since $b$ does not have an immediate predecessor, there is some $j'$ such that $j < j' < b$. So all neighborhoods of $b$ intersect $(a, b)$, meaning $b$ is a limit point of $(a, b)$.

Thus equality holds if and only if $a$ and $b$ have no immediate successor and predecessor respectively.

#6. (c)

In the space $\mathbb{R}$, if we let $A_\alpha = \{ q \}$ for all integers $q \in \mathbb{Q}$ then each $\overline{A_\alpha} = A_\alpha$ since $A_\alpha$ is a one-point set. But $\bigcup A_\alpha = \mathbb{Q}$, whose closure is $\mathbb{R}$. Thus $\bigcup \overline{A_\alpha} = \mathbb{Q}$ is smaller than $\overline{\bigcup A_\alpha} = \mathbb{R}$.

#10. Show that every order topology is Hausdorff.

Suppose $X$ is simply ordered, with the order topology. Suppose $x \neq y$ and without loss of generality, suppose $x < y$. Note that open rays are open in the order topology. There are two cases to consider:

(i) There is some $u \in X$ such that $x < u < y$. Then the open rays $(-\infty, u)$ and $(u, +\infty)$ are neighborhoods of $x$ and $y$ respectively that are disjoint.

(ii) There is no $u \in X$ such that $x < u < y$. Then the open rays $(-\infty, y)$ and $(x, +\infty)$ are neighborhoods of $x$ and $y$ respectively, and they are disjoint by the nonexistence of $u$.