Exercise 5. Let $X$ be a topological space. Show the following conditions hold:
a. The empty set and $X$ are closed.
Proof. $X=\emptyset^c$ is closed since $\emptyset$ is open and $\emptyset=X^c$ is closed since $X$ is open.
b. Arbitrary intersections of closed sets are closed.
Proof. Let $\cap_{\alpha \in \Delta} A_\alpha$ be an arbitrary intersection of closed sets. Then $\left(\cap_{\alpha \in \Delta} A_\alpha\right)^c=$ $\cup_{\alpha \in \Delta} A_\alpha^c$ is an arbitrary union of open sets and hence open. $\therefore \cap_{\alpha \in \Delta} A_\alpha$ is closed.
c. Finite unions of closed sets are closed.
Proof. Let $\cup_{i=1}^n A_i$ be a finite union of closed sets. Then $\left(\cup_{i=1}^n A_i\right)^c=\cap_{i=1}^n A_i^c$ is a finite intersection of open sets and hence open. $\therefore \cup_{i=1}^n A_i$ is closed.

a. The empty set and $X$ are closed.

Proof. We can observe that $X$ is the complement of the empty set, and since the empty set is open by definition, its complement $X$ is closed. Similarly, the empty set is the complement of $X$, and since $X$ is open by definition, its complement, the empty set, is closed. Therefore, both the empty set and $X$ are closed.

b. Arbitrary intersections of closed sets are closed.

Proof. Let ${A_\alpha}_{\alpha \in \Delta}$ be an arbitrary collection of closed sets, where $\Delta$ is an index set. We want to show that $\bigcap_{\alpha \in \Delta} A_\alpha$ is closed.

The complement of $\bigcap_{\alpha \in \Delta} A_\alpha$ is given by:
$$\left(\bigcap_{\alpha \in \Delta} A_\alpha\right)^c = \bigcup_{\alpha \in \Delta} A_\alpha^c.$$

Since each $A_\alpha$ is closed, $A_\alpha^c$ is open for every $\alpha \in \Delta$. Thus, $\bigcup_{\alpha \in \Delta} A_\alpha^c$ is a union of open sets, which means it is open.

Therefore, the complement of $\bigcap_{\alpha \in \Delta} A_\alpha$ is open, implying that $\bigcap_{\alpha \in \Delta} A_\alpha$ is closed.

c. Finite unions of closed sets are closed.

Proof. Let $A_1, A_2, \ldots, A_n$ be a finite collection of closed sets. We want to show that $\bigcup_{i=1}^n A_i$ is closed.

The complement of $\bigcup_{i=1}^n A_i$ is given by:
$$\left(\bigcup_{i=1}^n A_i\right)^c = \bigcap_{i=1}^n A_i^c.$$

Since each $A_i$ is closed, $A_i^c$ is open for $1 \leq i \leq n$. Thus, $\bigcap_{i=1}^n A_i^c$ is an intersection of open sets, which means it is open.

Therefore, the complement of $\bigcup_{i=1}^n A_i$ is open, implying that $\bigcup_{i=1}^n A_i$ is closed.

Exercise 7. Let $\tau$ and $\tau^{\prime}$ be two topologies on a set $X$ and let $i:\left(X, \tau^{\prime}\right) \rightarrow(X, \tau)$ be the identity map.
A) $\tau^{\prime}$ is finer than $\tau \Leftrightarrow i$ is continuous.
B) $\tau^{\prime}=\tau \Leftrightarrow i$ is a homeomorphism.
Proof. A) First assume $\tau^{\prime}$ is finer than $\tau$. Let $S$ be a set open in $(X, \tau)$. Then $S$ is open in $\left(X, \tau^{\prime}\right)$. Since $i:\left(X, \tau^{\prime}\right) \rightarrow(X, \tau)$ is the identity map, we have that $i^{-1}(S)=S$. Then $i^{-1}(S)$ is open in $\left(X, \tau^{\prime}\right)$. Therefore $i$ is continuous.

Next, let $i$ be continuous. Let $O$ be an open set in $\tau$. Then since $i$ is continuous, $i^{-1}(O)$ is open in $\left(X, \tau^{\prime}\right)$. Since $i^{-1}(O)=O, \tau^{\prime}$ is finer than $\tau$.

Proof. B) Assume $\tau=\tau^{\prime}$. Then $\tau^{\prime}$ is finer than $\tau$. By part A, $i$ is continuous. Since $\tau$ is also finer than $\tau^{\prime}, i^{-1}$ is also continuous. It is clear that $i$ is both one-to-one and onto. Then, $i$ is a homeomorphism.

Now, assume that $i$ is a homeomorphism. Then $i$ is continuous, so by part $\mathrm{A}, \tau^{\prime}$ is finer than $\tau . i^{-1}$ is also continuous, so $\tau$ is finer than $\tau^{\prime}$. Then, we must have $\tau=\tau^{\prime}$.

A) First, assume that $\tau’$ is finer than $\tau$. Let $S$ be a set open in $(X, \tau)$. Since $\tau’$ is finer than $\tau$, $S$ is also open in $(X, \tau’)$. Now, the identity map $i: (X, \tau’) \rightarrow (X, \tau)$ is defined as $i(x) = x$ for all $x \in X$. The preimage of $S$ under $i$ is given by $i^{-1}(S) = {x \in X : i(x) \in S} = {x \in X : x \in S} = S$. Since $S$ is open in $(X, \tau’)$, its preimage $S = i^{-1}(S)$ is open in $(X, \tau’)$. Therefore, $i$ is continuous.

Conversely, let $i$ be a continuous map. Now, let $O$ be an open set in $\tau$. Applying the continuity of $i$, we have $i^{-1}(O)$ is open in $\tau’$. However, since $i^{-1}(O) = O$ (as $i$ is the identity map), we conclude that $O$ is also open in $\tau’$. Thus, $\tau’$ is finer than $\tau$.

B) Assume $\tau = \tau’$. In this case, since $\tau’$ is finer than $\tau$ and $\tau$ is finer than $\tau’$ (as they are equal), we can use part A to conclude that $i$ is continuous. Moreover, since $\tau$ and $\tau’$ are equal, the preimage of any set under $i$ remains the same set. Thus, $i^{-1}(S) = S$ for all subsets $S$ of $X$. Now, if $i$ is a homeomorphism, it implies that $i$ is bijective (one-to-one and onto), and its inverse $i^{-1}$ is also continuous. This implies that $i^{-1}(S)$ is open in $\tau’$ for any set $S$ that is open in $\tau$. But since $i^{-1}(S) = S$ for all $S$, it means that every set that is open in $\tau$ is also open in $\tau’$. Therefore, $\tau$ and $\tau’$ are equal, $\tau = \tau’$.

Conversely, assume that $i$ is a homeomorphism. This implies that $i$ is continuous, bijective, and its inverse $i^{-1}$ is also continuous. From part A, we know that $i$ being continuous implies that $\tau’$ is finer than $\tau$. However, since $i^{-1}$ is also continuous, the same argument shows that $\tau$ is finer than $\tau’$. Therefore, we conclude that $\tau = \tau’$.

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