Pick some open cover of $X$ under $T$. This is also an open cover of $X$ under $T'$. Since $X$ is compact under $T'$, there exists a finite subcover of the open cover that covers $X$. This is also a finite subcover in $T$. Thus, $X$ is compact under $T$.

Assume, without loss of generality, that $T$ is a strict subset of $T'$ This implies that there is some open $U' \ \in \ T'$ such that $U' \ \notin \ T$. Now, pick some point $x \ \in \ U'$. Since $X \ - \ U'$ is closed under $T'$, it is compact under $T'$. It follows from the previous lemma that it is compact under $T$. Since $X$ is Hausdorff, $X \ - \ U'$ is closed under $T$, which implies that $U'$ is open under $T$, a contradiction. Thus, $T$ cannot be a strict subset of $T'$. A similar argument shows that $T'$ cannot be a strict subset of $T$. It follows that either $T \ = \ T'$ or $T$ and $T'$ are not comparable.

Consider the collection of compact subspaces $\{X_1, \ X_2, \ ..., \ X_n\}$. Consider an open cover $\mathcal{U}$ of their union. For each $X_j$, it follows that $\mathcal{U}$ is an open cover of $X_j$, so we find a finite subcover, $\mathcal{U}_j$. The union: $$U \ = \ \displaystyle\bigcup_{j} \mathcal{U}_j$$ will be a finite subcover of the union of all $X_j$, so the union is compact.

Take $Y$ to be a compact subspace of some metric space, $X$. Let us consider the open cover of $Y$, where each point $y \ \in \ Y$ is contained in the $\epsilon$-ball centred at $y$. Since $Y$ is compact, it follows that a finite subcollection of $N$ of these balls, which we call $\mathcal{B}$, cover the subspace: $$\mathcal{B} \ = \ \{B(y_1, \epsilon), \ B(y_2, \ \epsilon), \ ..., \ B(y_N, \ \epsilon)\}$$ Now, let us pick two points $x$ and $z$ of the metric space. Clearly, $x$ and $y$ must be contained in $\epsilon$-balls centred at some $y_x$ and some $y_z$. It follows from the triangle inequality that: $$d(x, \ z) \ < \ d(x, \ y_x) \ + \ d(y_x, \ y_z) \ + \ d(y_z \ + \ z) \ < \ 2\epsilon \ + \ d(y_x, \ y_z)$$ Since $N$ is finite, the set of all $d(y_x, \ y_z)$ is finite, and thus has a maximal element, $\alpha$. It follows that for any two points $x$ and $z$ of $Y$, we have $d(x, \ z) \ < \ 2\epsilon \ + \ \alpha$. This implies that each subspace must be bounded.

This is a direct result of Theorem 26.3. Every metric space is Hausdorff, thus, every compact subspace is closed.

We choose some point $a \ \in \ A$. Now, for each point $b \ \in \ B$, there exists open sets $U_{a}^{b}$ and $V_{a}^{b}$ around $a$ and $b$ that are disjoint. Since $b$ is compact, a finite union of such sets results in some $V_a$ which covers $B$, while the finite intersection of their counterparts around $a$ form a disjoint open set $U_a$, so $V_a \ \cap \ U_a \ = \ \emptyset$: $$V_a \ = \ V_{a}^{b_1} \ \cup \ ... \ \cup \ V_{a}^{b_n}$$ $$U_a \ = \ U_{a}^{b_1} \ \cap \ ... \ \cap \ U_{a}^{b_n}$$ Now, consider the set $\{U_a\}$ for all $a$. Since $A$ is compact, there is a finite subcover of this cover. The union of this finite subcover results in an open set $U$ that covers $A$. Taking the finite intersection of all corresponding $V_a$ gives us a disjoint open set $V$ that contains $B$.

Let us take $C$ to be closed in $X$. Since $X$ is compact, $C$ is also compact. It follows that $f(C)$ is compact. Since $Y$ is Hausdorff, $f(C)$ is closed.

Choose some closed set $C$ of $X \ \times \ Y$. Clearly, $X \ \times \ Y \ - \ C$ is open. By the tube lemma, for all $x_0 \ \times \ Y$ contained in $N$, it follows that $N$ contains a tube around $x_0 \ \times \ Y$. The union of all such tubes projected onto their first coordinate is an open set of $X$. The complement of this set in $X$ is $\pi_1(C)$. This implies that $\pi_1(C)$ is closed in $X$.

Assume that $f$ is continuous. We also know that $Y$ is compact and Hausdorff. We wish to show that the complement of $G_f$ is open. Pick some point $x \ \times \ y$ of the complement. Consider the two points $y$ and $f(x)$ of $Y$. Since $Y$ is Hausdorff, we know that there exists disjoint open sets $U$ and $V$ containing $y$ and $f(x)$ respectively. Since $f$ is continuous, we know that the set $f^{-1}(V)$ is open as well. Clearly, this set contains $x$, as $f(x) \ \in \ V$. Consider the set $A \ = \ f^{-1}(V) \ \times \ U$. Assume that there exists some $x \ \times \ f(x)$ in $A$. This implies that $x \ \in \ f^{-1}(V)$ and $f(x) \ \in \ U$. But from the first condition, we have $f(x) \ \in \ V$, a clear contradiction as $V$ and $U$ are disjoint. Thus, the open set $A$ around $x \ \times \ y$ is disjoint from $G_f$. It follows that the complement of $G_f$ is open, and $G_f$ is closed.

Conversely, assume that $G_f$ is closed in $X \ \times \ Y$. Consider some closed $V$ of $Y$. It follows that $G_f \ \cap \ X \ \times \ V$ is closed in $X \ \times \ Y$. Note that this set is exactly the set of all $x \ \times \ f(x)$ such that $f(x) \ \in \ V$. From the previous lemma, it follows that $\pi_{1} : X \ \times \ Y \ \rightarrow \ X$ is closed. It follows that: $$\pi_1 ( G_f \ \cap \ X \ \times \ V) \ = \ \{x \ | \ f(x) \ \in \ V\} \ = \ f^{-1}(V)$$ is closed. This is implies that $f$ is continuous.

For each $\epsilon$ > $0$, there is some $N$ such that for all $n \ \geq \ N$: $$| f(x) \ - \ f_n(x) | \ < \ \epsilon/3$$ For each $x$, we denote this value of $N$ by $N_x$. Now, for each $x$, let $B_{\epsilon/3}(f(x))$ be the $\epsilon/3$ ball around $f(x)$, and let $B_{\epsilon/3}(f_{N_x}(x))$ be the $\epsilon/3$ ball around $f_{N_x}(x)$.

Since both functions are continuous, we can choose neighbourhoods $U_x$ and $V_x$ of $x$, such that $f(U_x) \subset B_{\epsilon/3}(f(x))$ and $f(V_x) \subset B_{\epsilon/3}(f_{N_x}(x))$. We let $W_x = U_x \cap V_x$. Clearly, this open set will be non-empty, as it contains $x$.

Since $X$ is compact, we choose some finite subcover $W_{x_1}, \ ..., \ W_{x_m}$. Consider some $y \in X$. It must be true that $y \in W_{x_k}$, for some $k$. We then note that $f_{N_{x_k}}(y) \in B_{\epsilon/3}(f_{N_{x_k}}(x_k))$ and $f(y) \in B_{\epsilon/3}(f(x))$.

Thus:

$$|f_{N_{x_k}}(y) - f_{N_{x_k}}(x_k)| < \epsilon/3$$ $$|f_{N_{x_k}}(x_k) - f(x)| < \epsilon/3$$ $$|f(x) - f(y)| < \epsilon/3$$ so it follows that $|f_{N_{x_k}}(y) - f(y)| < \epsilon$. Since the functions are increasing montone, we must have $f_n(y) \leq f(y)$, for all $n$, or else the sequence would not converge. Thus, for any $n \geq N_{x_k}$, we must also have $|f_n(y) - f(y)| < \epsilon$.

Finally, we let $N = \max \{N_{x_k} \ | \ 1 \leq k \leq m\}$. For this $N$, we will have, from above, for all $n \geq N$, then $|f_n(x) - f(x)| < \epsilon$, for all $x$. Thus, $f$ converges uniformly.

Let us assume that there exists a separation $C \ \cup \ D$ of $Y$. Clearly, $Y$ is closed in $X$. Since $C \ = \ Y \ - \ D$ and $D \ = \ Y \ - \ C$, both $C$ and $D$ are closed in $Y$. It follows that they are both closed in $X$. Since all closed subspaces of a compact set are also compact, it follows from Problem 5 that we can choose disjoint open set $U$ and $V$ around $C$ and $D$. Now, consider the following intersection: $$Z \ = \ \displaystyle\bigcap_{A \in \mathcal{A}} (A \ - \ (U \ \cup \ V))$$ Clearly $A \ - \ (U \ \cup \ V)$ is closed in $A$. Since each $A$ is closed in $X$, it follows that $A \ - \ (U \ \cup \ V)$ is closed in $X$. All of the sets $A \ \in \ \mathcal{A}$ are ordered by proper inclusion. In addition, each $A \ - \ (U \ \cup \ V)$ must be non-empty, or else $A \ \cap \ U$ and $A \ \cap \ V$ would form a separation of $A$, which contradicts the fact that each $A$ is connected. It follows that the collection of closed set $A \ - \ (U \ \cup \ V)$ for $A \ \in \ \mathcal{A}$ has the finite intersection property. Since $X$ is compact, it follows that $Z$ must be non-empty. This is clearly a contradiction, as: $$Y \ \subset \ U \ \cup \ V$$ This would imply that we have "deleted" all of the points that the sets $A \ \in \ \mathcal{A}$ have in common, and yet, their intersection is still non-empty, which can't possibly be the case. Thus, it must be true that no separation $C \ \cup \ D$ exists, and $Y$ is connected.

Consider the open set $U$ of $X$. Consider some $p^{-1}(\{y\}) \ \subset \ U$. It follows that $y \ \in \ p(U)$. Now, since $p$ is a closed map, we know that the set $W \ = \ Y \ - \ p(X \ - \ U)$ is open. We assert that $y \ \in \ W$. Assume that $y$ wasn't in $W$. It follows that $y \ \in \ p(X \ - \ U)$. This implies that there exists some point $x$ of $X \ - \ U$ such that $p(x) \ = \ y$, as $p$ is surjective. This would imply that $x \ \in \ p^{-1}(\{y\})$, which would mean that $p^{-1}(\{y\})$ intersects $X \ - \ U$, a contradiction. It is easy to see that $p^{-1}(W) \ \subset \ U$. Thus, for some open $U$ containing $p^{-1}(\{y\})$, there exists a neighbourhood $W$ of $y$ such that $p^{-1}(W)$ is in $U$.

Now, consider some open cover of $X$, which we call $\mathcal{U}$. Consider the collection of all sets of the form $p^{-1}(\{y\})$. The union of all such sets, for all $y \ \in \ Y$ will equal all of $X$. For each $p^{-1}(\{y\})$ consider all of the open sets of $\mathcal{U}$ that intersect it, which we call $\mathcal{U}_y$. Since $p^{-1}(\{y\})$ is compact, it follows that a finite collection of these sets will cover it: $$p^{-1}(\{y\}) \ \subset \ U_y \ = \ U_1 \ \cup \ U_2 \ \cup \ ... \ \cup \ U_n \ \ \ \ \ \ \ U_j \ \in \ \mathcal{U}_y$$ Now, for each $p^{-1}(\{y\})$, let us consider the open set $W$ around $y$ such that $p^{-1}(W) \ \subset \ U_y$. Since $Y$ is compact, a finite number of such $W$ sets cover $Y$: $$W_1 \ \cup \ W_2 \ \cup \ ... \ \cup \ W_n \ = \ Y$$ Now comes the interesting part: for each $W_j$, we take the inverse image $p^{-1}(W_j)$. We know that each $p^{-1}(W_j)$ is contained in some $U_y$, which in turn is equal to the union of a finite number of open sets in the cover $\mathcal{U}$. Thus: $$p^{-1}(W_1) \ \cup \ p^{-1}(W_2) \ \cup \ ... \ p^{-1}(W_n) \ = \ p^{-1}(Y) \ = \ X$$ Is equal to the union of a finite number of open sets, with each open set in the union being a finite union of open sets of $\mathcal{U}$. A finite collection of finite sets is finite. Thus, there exists a finite subcover of $X$ with respect to $\mathcal{U}$. It follows that $X$ is compact.