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p. 83: 3 [adapted from latex HW of Colin McKinney]
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Show that the collection $\mathcal{T}_c$ given in Example 4 of $\S12$ is a
topology on the set $X$. \\ \\ \noindent $\mathcal{T}_c := \{ U | X - U $ is countable or all of $X
\}$.  First, we must ensure that $X$ and $\emptyset$ are elements of $\mathcal{T}_c$.  For $U = \emptyset$, $X -
\emptyset = X$.  Thus, $\emptyset \in \mathcal{T}_c$.  For $U = X$, $X - X = \emptyset$.  The null set is
finite, and so it is also countable.  Hence $X \in \mathcal{T}_c$. 
\\ \\ \noindent We must also show that
$\mathcal{T}_c$ is closed under arbitrary unions.  Take $U_\alpha \in \mathcal{T}_c$. Since $X-U_{\alpha}$ can be
either countable or all of $X$, we must consider two cases. \\ \\ \noindent \begin{bf}Case 1 \end{bf}:  There
exists an $\alpha_0$ such that $X - U_{\alpha_0}$ countable.  $X - \cup U_{\alpha} = \cap (X - U_{\alpha})$, and
since $ X - U_{\alpha_0} $ is countable, and $\cap (X - U_{\alpha}) \subset X - U_{\alpha_0}$, it follows that
$\cap (X - U_{\alpha})$ is countable.  Hence $ X - \cup U_{\alpha}$ is countable, and so in this case $\cup
U_{\alpha} \in \mathcal{T}_c$. \\ \\ \noindent
\begin{bf}Case 2 \end{bf}: $ X - U_{\alpha}$ is all of X for all $\alpha$.  Hence $U_{\alpha} = \emptyset$ for all 
$\alpha$. It follows that $\cup U_{\alpha} = \emptyset \in \mathcal{T}_c$ 
\\ \\
Hence $\mathcal{T}_c$ is closed under arbitrary unions.  
\\ \\ \noindent Lastly we must show that $\mathcal{T}_c$
is closed under finite intersections.  Take $U_i \in \mathcal{T}_c$. \\ \\ \noindent
\begin{bf}Case 1 \end{bf}: $X - U_i$ countable for all $i$.  A finite union of countable sets is also countable.  
By Problem 1.0 of Homework 1, we may write \begin{equation}X - \bigcap_{i=1}^{n}U_{i} =
\bigcup_{i=1}^{n} (X - U_{i}) \end{equation} \noindent   Since $\cup(X -
U_{i})$ is countable, the left hand side of (1) is also countable.  Thus, 
$\cup_{i=1}^{n}U_{i} \in
\mathcal{T}_c$ in this case. 
\\ \\ \noindent \begin{bf}Case 2
\end{bf}: $X - U_{i_0}$ is all of $X$ for some $i_0 \in \{1, ..., 1 \}$.  Hence $U_{i_0} = \emptyset$, and 
$\cap_{i=1}^{n} 
U_{i} =
\emptyset \in \mathcal{T}_c$. 
\\ \\
 Thus, $\mathcal{T}_c$ 
is
closed under finite intersections.  $\mathcal{T}_c$ has thus satisfied all the requirements to be a topology. \\ \\
\\ \\ \noindent  Is $\mathcal{T}_{\infty} := \{U | X - U $ is infinite, empty, or all of $X
\}$ a topology on $X$?
\\ \\
If $X$ is infinite, $U_x = \{x\} \in \mathcal{T}_{\infty} $, where $x$ is any element of $X$ since
$X - U_x$ would remain infinite.  However, let us define an indexing set $Y := X - \{x_1, x_2\}$, where $x_1$ and
$x_2$ are two distinct elements of $X$.  Hence \begin{equation} X - \bigcup_{x \in Y} U_x = \{x_1, x_2\}. 
\end{equation}
Since $\{x_1, x_2\}$ is finite, $X - \cup_{x \in Y} U_{x}$ is finite.  Hence $\cup_{x \in Y} U_{x}$ is not an 
element
of $\mathcal{T}_{\infty}$.  Thus when $X$ is infinite, $\mathcal{T}_{\infty}$ is not a topology since it is not 
closed under 
unions. \\ \\ \noindent 
 \\ \\ \noindent 
Suppose $X$ is finite.  
Note $\emptyset$ and $X$ are elements of
$\mathcal{T}_{\infty}$.  If $U = \emptyset$, then $X - U = X$, so $\emptyset \in \mathcal{T}_{\infty}$.  If $U =
X$, then $X - X = \emptyset$, and so $X \in \mathcal{T}_{\infty}$.  
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Let $U \in \mathcal{T}_{\infty}$.  Since $X$ is finite, $X - U$ is not infinite.  Thus $X - 
U = \emptyset$ or $X - U = X$.  Thus, $U = X$ or $U = \emptyset$.  Hence $\mathcal{T}_{\infty}$ is the indiscrete 
topology and is thus a topology.
\\ \\ \noindent 



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