Exercise 17.1

Question

Let $\mathcal{C}$ be a collection of subsets of the set $X$.
  Suppose that $\varnothing$ and $X$ are in $C$, and that finite unions and arbitrary intersections of elements of $\mathcal{C}$ are in $\mathcal{C}$.
  Show that the collection
  \begin{gather*}
    \mathcal{T} = \left\{X - C \mid C \in \mathcal{C}\right\}
  \end{gather*}
  is a topology on $X$.

Dan Whitman · Accepted Answer

\begin{proof}
  First, clearly $\varnothing$ and $X$ are in $\mathcal{T}$ since $\varnothing = X - X$ and $X = X - \varnothing$ and both $X$ and $\varnothing$ are in $\mathcal{C}$.
  This shows the first defining property of a topology.

Now consider an arbitrary sub-collection $\mathcal{A}$ of $\mathcal{T}$.
  Then, for each $A \in \mathcal{A}$, we have that $A = X - B$ for some $B \in \mathcal{C}$ since also $A \in \mathcal{T}$.
  So let $\mathcal{B} = \left\{B \in \mathcal{C} \mid X - B \in \mathcal{A}\right\}$.
  Then we have that
  \begin{gather*}
    \bigcup \mathcal{A} = \bigcup_{A \in \mathcal{A}} A = \bigcup_{B \in \mathcal{B}} (X - B) = X - \bigcap_{B \in \mathcal{B}} B = X - \bigcap \mathcal{B}
  \end{gather*}
  by DeMorgan's law.
  By the definition of $\mathcal{C}$ we have that $\bigcap \mathcal{B} \in \mathcal{C}$ since it is an arbitrary intersection of elements of $\mathcal{C}$.
  It then follows that $\bigcup \mathcal{A} = X - \bigcap \mathcal{B}$ is in $\mathcal{T}$ by definition.
  This shows the second defining property of a topology.

Lastly, suppose that $\mathcal{A}$ is a nonempty finite sub-collection of $\mathcal{T}$, which of course can be expressed as $\mathcal{A} = \left\{A_k \mid k \in \left\{1, \ldots, n\right\}\right\}$ for some positive integer $n$.
  Then, again we have that that $A_k = X - B_k$ for some $B_k \in \mathcal{C}$ for all $k \in \left\{1, \ldots, n\right\}$ since $A_k \in \mathcal{T}$.
  Then we have
  \begin{gather*}
    \bigcap \mathcal{A} = \bigcap_{k=1}^n A_k = \bigcap_{k=1}^n (X - B_k) = X - \bigcup_{k=1}^n B_k
  \end{gather*}
  by DeMorgan's law.
  Then, clearly, $\bigcup_{k=1}^n B_k$ is in $\mathcal{C}$ by definition since it is a finite union of elements of $\mathcal{C}$.
  It then follows that $\bigcap \mathcal{A} = X - \bigcup_{k=1}^n B_k$ is in $\mathcal{T}$ by definition.
  Since $\mathcal{A}$ was an arbitrary finite sub-collection, this shows the third defining property of a topology.
  Hence $\mathcal{T}$ is a topology by definition.
\end{proof}

Exercise 17.1

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Exercise 17.1

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