answered #1, #2(a), #3, #4 (except last part), #5
wrote up #1, #2(a), #5
(a) Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on the set $X$; suppose $\mathcal{T} \subset \mathcal{T}'$. What does compactness of $X$ under one of these topologies imply about the other?
If $X$ is compact under $\mathcal{T}$ we do not have any implication for $\mathcal{T}'$.
For example, let $X = [0, 1]$ and let $\mathcal{T}$ be the order topology and $\mathcal{T}'$ be the discrete topology. $X$ is compact under $\mathcal{T}$ (as we'll prove later) and but not under $\mathcal{T}'$; in particular, the open covering
$$ \mathcal{C} = \left\{ \{ r \} \mid r \in [0, 1] \right\} $$
made up of single point sets has no finite sub-covering (in fact, no strict subset of it is a covering at all!).
In contrast, the topology $\mathcal{T}$ itself can be viewed as a finer topology and it is compact.
So some finer topologies than $\mathcal{T}$ are compact while others aren't, so we have no conclusion about an arbitrary $\mathcal{T}'$.
In contrast, if $X$ is compact under $\mathcal{T}'$, then it is also compact under $\mathcal{T}$.
Let $\mathcal{C}$ be an open covering of $X$ under $\mathcal{T}$. Then it is also an open covering of $X$ under $\mathcal{T}'$. Then there is a finite sub-covering in $\mathcal{T}'$ since that topology is compact. But that sub-covering is also a sub-covering in $\mathcal{T}$! So $X$ is also compact under $\mathcal{T}$.
(b) Show that if $X$ is compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T}'$, then either $\mathcal{T}$ and $\mathcal{T}'$ are equal or they are not comparable.
Without loss of generality, suppose $\mathcal{T} \subset \mathcal{T}'$. Consider the identity function
$$ f: X \text{ (under } \mathcal{T}' \text{)} \rightarrow X \text{ (under } \mathcal{T} \text{)}, \,f(x) = x. $$
Note that $f$ is bijective, and it is continuous by the relationship between $\mathcal{T}$ and $\mathcal{T}'$.
And since both topologies are compact and Hausdorff, by theorem 26.6, $f$ is a homeomorphism. Then $f(U)$ is open if and only if $U$ is open. In other words, $U$ is in $\mathcal{T}$ if and only if it is in $\mathcal{T}'$. So the two topologies are equal.
Thus, if the two topologies are comparable, they are equal. So they are either equal or not comparable at all.
(a) Show that in the finite complement topology on $\mathbb{R}$, every subspace is compact.
Consider any $S \subset \mathbb{R}$. Suppose $\mathcal{C}$ is an open covering of $S$ in $\mathbb{R}$.
Note that every nonempty $U \in \mathcal{C}$ has a finite complement in $\mathbb{R}$.
Take any nonempty $U_0 \in \mathcal{C}$ (if $\mathcal{C}$ only contains the empty set then it is already finite). Then $U_0$ contains all but finitely many points of $\mathbb{R}$. Then $U_0$ also contains all but finitely many points of $S$, say $\{ s_1, \ldots, s_n \}$.
For each $s_i$, pick some $U_i \in \mathcal{C}$ such that $s_i \in U_i$. This is possible since $\mathcal{C}$ covers $S$. Then $\bigcup_{i=0}^n U_i$ is a finite sub-covering for $S$.
Thus $S$ is compact.
[stuck on 2(b)]
Let $A$ and $B$ be disjoint compact subspaces of the Hausdorff space $X$. Show that there exist disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively.
Let $a \in A$. By lemma 26.4, there are disjoint open sets $U_a$ and $V_a$ that contain $a$ and $B$ respectively. Then define the collection:
$$ \mathcal{U} = \{ U_a \mid a \in A \} $$
This collection is an open covering of $A$, hence there is a finite sub-covering
$$ \mathcal{U}F = \{ U{a_i} \mid i = 1, \ldots, N \} $$
We define the corresponding collection $\mathcal{V}_F$ as
$$ \mathcal{V}F = \{ V{a_i} \mid i = 1, \ldots, N \} $$
Finally, define $U$ and $V$ as
$$ U = \bigcup_{i=1}^N U_{a_i} \, \text{ and } V = \bigcap_{i=1}^N V_{a_i} $$
respectively.
By construction these are disjoint open sets containing $A$ and $B$ respectively.