Exercise 13.3

Question

Show that the collection $\mathcal{T}_{\alpha}$ given in Example~4 of \S 12 is a topology on $X$.
Is the collection
\begin{gather*}
\mathcal{T}_{\infty} = \left\{U \mid X - U 	ext{ is infinite or empty or all of } Xight\}
\end{gather*}
a topology on $X$?

Amit Awasthi · Accepted Answer

Recall Example 12.4: Let $X$ be a set; let $\mathcal{T}_c$ be the
  collection of all subsets $U$ of $X$ such that $X \setminus U$ either is
  countable or is all of $X$. We claim this forms a topology on $X$; we will
  follow the numbering for the definition of a topology on p.~76. (1) $X
  \setminus \emptyset = X$ and $X \setminus X = \emptyset$ is countable; (2)
  $X \setminus \bigcup_{\alpha \in A} U_\alpha = \bigcap_{\alpha \in A} (X
  \setminus U_\alpha)$ is countable since it is an intersection of countable sets,
  unless every $U_\alpha = \emptyset$, in which case $X \setminus \bigcup_{\alpha \in
  A} U_\alpha = X$; (3) $X \setminus \bigcap_{\alpha \in A~	ext{finite}}
  U_\alpha = \bigcup_{\alpha \in A~	ext{finite}} (X \setminus U_\alpha)$ is
  countable since it is the finite union of countable sets, unless every
  $U_\alpha = \emptyset$, in which case $X \setminus
  \bigcap_{\alpha \in A~	ext{finite}} U_\alpha = X$.
  \par Now consider $\mathcal{T}_\infty$. It is not a topology, for if we let $X = [-1,1] \subseteq \mathbb{R}$, and let $U_1 = [-1,0)$ and $U_2 = (0,1]$, we see that both $U_1,U_2 \in \mathcal{T}_\infty$, but $X \setminus (U_1 \cup U_2) = \{0\}$, which is not infinite, and so $U_1 \cup U_2 
otin \mathcal{T}_\infty$.

Dan Whitman · Answer

Recall that $\mathcal{T}_{\alpha}$ from Example~12.4 is the set of all subsets $U$ of $X$ such that $X - U$ either is countable or is all of $X$.
First, we show that $\mathcal{T}_{\alpha}$ is a topology on $X$.
\begin{proof}
First, clearly $\varnothing \in \mathcal{T}_{\alpha}$ since $X - \varnothing = X$ is all of $X$.
Also $X \in \mathcal{T}_{\alpha}$ since $X - X = \varnothing$ is countable.
Now suppose that $\mathcal{A}$ is a subcollection of $\mathcal{T}_{\alpha}$ so that $X - U$ is countable (or all of $X$) for every $U \in \mathcal{A}$.
Then we have that
\begin{gather*}
X - \bigcup \mathcal{A} = X - \bigcup_{A \in \mathcal{A}} A = \bigcap_{A \in \mathcal{A}} (X - A)
\end{gather*}
is countable (or all of $X$) since every $X - A$ is countable (or all of $X$).
Therefore $\bigcup \mathcal{A} \in \mathcal{T}_{\alpha}$ by definition.

Now suppose that $U_1, \ldots, U_n$ are nonempty elements of $\mathcal{T}_{\alpha}$ so that $X - U_i$ is a countable subset of $X$ or $X$ itself for each $i \in \left\{1, \ldots, night\}$.
Then we have
\begin{gather*}
X - \bigcap_{i=1}^n U_i = \bigcup_{i=1}^n (X - U_i)
\end{gather*}
is a finite union of sets that are either countable subsets of $X$, or $X$ itself.
It then follows that the union is countable or $X$ itself so that $\bigcap_{i=1}^n U_i \in \mathcal{T}_{\alpha}$ by definition.
This completes the proof that $\mathcal{T}_{\alpha}$ is a topology on $X$.
\end{proof}

Now we claim that the collection $\mathcal{T}_{\infty}$ as defined above is not always a topology on $X$.
\begin{proof}
As a counterexample, let $X = {{\mathbb{Z}}_+}$ and suppose that $\mathcal{T}_{\infty}$ is a topology on $X$.
Clearly if $U$ is a finite subset of $X$, then $X - U$ is infinite since $X$ is infinite so that $U$ is open.
Now consider the subcollection
\begin{gather*}
\mathcal{A} = \left\{\left\{iight\} \mid i \in {{\mathbb{Z}}_+} 	ext{ and } i > 1ight\} = \left\{\left\{2ight\}, \left\{3ight\}, \ldotsight\} \,.
\end{gather*}
Then clearly we have that $\bigcup \mathcal{A} = \left\{2,3, \ldotsight\}$ so that $X - \bigcup \mathcal{A} = \left\{1ight\}$ is neither infinite, empty, nor all of $X$.
Therefore $\bigcup \mathcal{A}$ cannot be open, which violates property (2) of a topology.
So it must be that $\mathcal{T}_{\infty}$ is not a topology, which of course contradicts our supposition that it is!
\end{proof}

Exercise 13.3

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Exercise 13.3

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