1

My assignment is the following: Let $X$ be a topological space with topology $T$. Define \begin{align*} Q =\{ A \subset P(X) \mid \bigcup_{a \in A} a\in T\} \\ R = \{ A \subset P(X) \mid \bigcap_{a \in A}a \in T\} \end{align*}

Determine whether or not $Q$ and $R$ are topologies on $P(X)$.

I would appreciate any hints as to how to begin this problem. This is what I have done:

$\forall U_1, U_2,...\in Q$, let $\bigcup_{i \in I} U_i=U$ and let $U_i=\{u_1, u_2,...\}$. Since $U_i \in Q$, $\bigcap_{j \in J} u_j=u_i \in T$. $U \in Q$ if and only if $\bigcup_{j \in J} \bigcup_{i \in I} u_{i,j} \in T$ this is where I am getting stuck and would appreciate some help.

tom
  • 87

1 Answers1

3

For clarity, I will use different fonts to indicate different sorts of objects:

  • I will use standard lowercase letters (e.g.: $x,y,z$) to indicate elements of $X$.
  • I will use standard uppercase letters (e.g.: $A,B,C$) to indicate subsets of $X$--i.e.: elements of $\mathcal P(X).$
  • I will use uppercase calligraphic letters (e.g.: $\mathcal{M,N,O}$) to indicate sets of subsets of $X$--i.e.: subsets of $\mathcal P(X).$
  • I will use uppercase script letters to describe the two sets of sets of subsets of $X$--i.e.: sets of subsets of $\mathcal P(X)$--with which this problem is concerned.

So, rephrasing your problem with this alternate notation in mind, we are taking $X$ to be a set, $\mathcal T$ a topology on $X,$ and defining sets $$\mathscr Q:=\left\{\mathcal M\subseteq\mathcal P(X):\bigcup_{A\in\mathcal M}A\in\mathcal T\right\}$$ and $$\mathscr R:=\left\{\mathcal M\subseteq\mathcal P(X):\bigcap_{A\in\mathcal M}A\in\mathcal T\right\}$$

Set-theoretically speaking, $\bigcap_{A\in\emptyset}A$ isn't well-defined, so $\emptyset\notin\mathscr R.$ Thus, $\mathscr R$ is not a topology on $\mathcal P(X).$

Now, suppose that $X$ is the real line with the usual topology, and let $$\mathcal M:=\bigl\{\{1,x\}:x\in X\bigr\}$$ $$\mathcal N:=\bigl\{\{2,x\}:x\in X\bigr\}.$$ We can see that $$\bigcup_{A\in\mathcal M}A=\bigcup_{A\in\mathcal N}A=X,$$ which is (of course) open in the usual topology, and so $\mathcal{M,N}\in\mathscr Q.$ However, $$\mathcal M\cap\mathcal N=\bigl\{\{1,2\}\bigr\},$$ so $\bigcap_{A\in\mathcal M\cap\mathcal N}A=\{1,2\},$ which is not open in the usual topology. Thus, $\mathscr Q$ is not closed under finite intersections, and so is not a topology on $\mathcal P(X).$

A similar argument works for any topological space in which some two-element subset is non-open. If every two-element subset is open, then either the space has exactly two points, or the space has more or less than two points and one can prove that every one-element subset (and therefore every subset) is open. In the latter case, $\mathcal T$ is the discrete topology on $X,$ which will make $\mathscr Q$ a topology--namely, the discrete topology--on $\mathcal P(X).$ In the two-element case, one can also verify that only the discrete topology on $X$ will make $\mathscr Q$ a topology on $\mathcal P(X).$

Addendum: Now, the argument could be made that we should instead define our notation as follows:

For any $\mathcal M\subseteq\mathcal P(X),$ say that $$\bigcup_{A\in\mathcal M}A:=\left\{x\in X:x\in A\text{ for some }A\in\mathcal M\right\}$$ and $$\bigcap_{A\in\mathcal M}A:=\left\{x\in X:x\in A\text{ for all }A\in\mathcal M\right\}.$$

This is sensible enough. Since we're only considering unions and intersections of subsets of $X,$ we should insist that the results of such set operations also be subsets of $X.$ In such a case, we can say that $\bigcup_{A\in\emptyset}A=\emptyset$ and $\bigcap_{A\in\emptyset}A=X.$ (Why?)

Using this convention does not have any effect on the validity of the counterexample for $\mathscr Q$ given above, but it does mean we would now have that $\emptyset\in\mathscr R.$

However, even this convention need not make $\mathscr R$ a topology on $\mathcal P(X),$ as a rule. Again, take $X$ to be the real line under the usual topology, and let $$\mathcal M=\bigl\{\{0\},\{1\}\bigr\}$$ and $$\mathcal N=\bigl\{\{0\},\{2\}\bigr\}.$$ Now, $$\bigcap_{A\in\mathcal M}A=\bigcap_{A\in\mathcal N}A=\emptyset,$$ which is open in the usual topology, and so $\mathcal{M,N}\in\mathscr R.$ However, $$\mathcal M\cap\mathcal N=\bigl\{\{0\}\bigr\},$$ so $\bigcap_{A\in\mathcal M\cap\mathcal N}A=\{0\},$ which is not open in the usual topology. Thus, $\mathscr R$ is not closed under finite intersections, and so is not a topology on $\mathcal P(X).$

A similar argument works in any non-discrete topological space. However, if the space is discrete, then $\mathscr R$ will be a topology--in particular, the discrete topology--on $\mathcal P(X).$

Cameron Buie
  • 102,994