Prove that $\mathcal T$ is a topology

Question

Let a function $h:\mathcal P(X)\to\mathcal P(X)$ be defined by

$h(\emptyset)=\emptyset$

$h(A\cup B)=h(A)\cup h(B),\;\forall A,B\in\mathcal P(X)$

$h(A)\supseteq A,\;\forall A\in\mathcal P(X)$

$h\circ h=h$

Now setup $\mathcal T:=\{A^\complement\in\mathcal P(X): h(A)=A\}$. Prove that $\mathcal T$ is a topology on $X$.

Im stuck with this problem. I can show that $\emptyset,X\in\mathcal T$ and that the finite intersection of elements of $\mathcal T$ belong to $\mathcal T$.

But Im unable to prove that arbitrary union of elements of $\mathcal T$ belong to $\mathcal T$. I was playing around with the properties of $h$ but I cant conclude something about this last axiom to define a topology.

Some hint or solution will be appreciated. Thank you.

@J.-E.Pin yep, you are right. Arbitrary union... but I never used uncountable index. — Masacroso, Nov 09 '16 at 12:54
Notably, $h$ functions as a "closure" function over this topology — Ben Grossmann, Nov 09 '16 at 12:56
Choose $A_i^C=X\setminus A_i\in \mathcal{T}$ arbitrarily (here $i\in I$ for some index set). In order to show that $\cup_i A_i^C\in \mathcal{T}$, we must show that $h(\cap_i A_i)=\cap_i A_i$. Now since $A_i^C\in \mathcal{T}$, $h(A_i)=A_i$. Thus $\cap_i A_i=\cap_i h(A_i)$. Hence we need to show that $h$ goes trough intersections of elements of $\mathcal{T}$. — Mathematician 42, Nov 09 '16 at 13:06
And notably, "going through intersections" is a property of any function over $X$; the remaining bit of Math42's proof doesn't use any more properties of $h$. — Ben Grossmann, Nov 09 '16 at 13:11
@Omnomnomnom: Haha, absolutely right, how could I have forgotten to use that very basic fact. Well, there you go then :) — Mathematician 42, Nov 09 '16 at 13:14
@Omnomnomnom but the function $h$ is defined over $\mathcal P(X)$, not over $X$. If the function $h$ is induced from a function over $X$ the solution is clear, yes. — Masacroso, Nov 09 '16 at 13:14
Whoops. Well, then I guess it's not so straightforward afterall — Ben Grossmann, Nov 09 '16 at 13:16
@Omnomnomnom probably I must assume that the function is induced from the function $h$ over $X$. — Masacroso, Nov 09 '16 at 13:16
That doesn't sound right. Again, $h$ is really a stand-in for taking the closure of a set; set closures can't always be induced from a function over $X$. — Ben Grossmann, Nov 09 '16 at 13:18
Ha again, indeed, it's not that simple. We probably need to use that $h\circ h=h$ somehow, but I'm still trying to figure out how to do this. — Mathematician 42, Nov 09 '16 at 13:29
Brilliant question, this baffled me a couple of times and the final proof is completely different from what I thought it was going to be. It turns out to be easy and uses absolutely nothing. Once you figure out that $h$ is inclusion preserving (on all subsets of $X$ even) it becomes very straightforward. Where did you encounter this question? — Mathematician 42, Nov 09 '16 at 14:09
@Mathematician42 page 247 in Analysis I of Amann and Escher, exercise 7. — Masacroso, Nov 09 '16 at 22:18

score 4 · Accepted Answer · answered Nov 09 '16 at 14:01

We are first going to show that $h$ preserves inclusions. Suppose that $A\subset B$. This is equivalent to saying that there exists an $C\in \mathcal{P}(X)$ such that $A\cup C=B$. Applying $h$ to this equality, we find that $h(A\cup C)=h(A)\cup h(C)=h(B)$, here we used that $h$ goes through finite unions. Since $h(A)\cup h(C)=h(B)$, we have that $h(A)\subset h(B)$. Thus we showed that $A\subset B \Rightarrow h(A)\subset h(B)$. (The converse is not necessarily true).

Now choose $A_i^C\in \mathcal{T}$ arbitrarily. Then $\cap_i A_i\subset h(\cap_i A_i)$ by the third property of $h$. Notice that $\cap_i A_i\subset A_i$, applying $h$ to this inclusion, we get that $$h(\cap_iA_i)\subset h(A_i)=A_i$$ holds for all $i$. Here we used that $h(A_i)=A_i$ since $A_i^C\in \mathcal{T}$. Since the above property holds for all $i$, we conclude that $h(\cap_i A_i)\subset \cap_i A_i$. Thus $h(\cap_i A_i)=\cap_i A_i$, which we needed to show by my comment above. Notice that we didn't use that $h\circ h=h$.

The fact that $h\circ h=h$ is not used is interesting. If I'm not mistaken, without that condition, the closure operation of that topology would be $\lim_{n\to\infty} h^{\circ n}$. Note that condition (3) ensures that this limit exists. — celtschk, Nov 14 '16 at 22:44

score 0 · Answer 2 · answered Nov 12 '16 at 22:02

To show that any union of subsets of $T$ belongs to $T,$ it suffices to show that if $Y\subset P(X)$ and $\forall B\in Y\;(h(B)=B)$ then $h(\cap Y)=\cap Y.$

(1).If $C\subset D\subset X$ then $h(C)\subset h(D)$ because $h(D)=h( C\cup (D$ \ $C))=h(C)\cup h(D$ \ $C).$

(2). If $Y\subset P(X)$ and $\forall B\in Y\;(B=h(B))$ then by (1), $$\forall B\in Y\; (h(\cap Y)\subset h(B)).$$ $$ \text { Therefore }\quad \cap Y\subset h(\cap Y)\subset \cap_{B\in Y}h(B)=\cap_{B\in Y}B=\cap Y.$$

MarnixKlooster ReinstateMonica · Answer 3 · 2020-01-31T20:56:39.150

Here is Mathematician 42's proof, but written down in a different form, with two goals. First, I wanted to convince myself that OP's property 4, $\;h \circ h = h\;$, is indeed not used, and I found it difficult to understand some details of that earlier answer. And second, to present the proof in a different order, making it clearer that '$\;h\;$ preserves $\;\subseteq\;$' is indeed a reasonable thing to look at.$% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\T}{\mathcal T} \newcommand{\c}{\complement} %$

First we note that, by the definition of $\;\T\;$, $$ \tag{0} B \in \T \;\equiv\; h(B^\c) = B^\c $$ In other words, this topology are the sets whose complements remain unchanged under $\;h\;$.

So, given an arbitrary $\;V \subseteq \T\;$, we calculate when its union is in $\;\T\;$: $$\calc \cup V \in \T \op\equiv\hint{expand definition of $\;\T\;$ using $\Ref{0}$} h((\cup V)^\c) \;=\; (\cup V)^\c \op\equiv\hint{OP's property 3 -- to reduce our proof burden} h((\cup V)^\c) \;\subseteq\; (\cup V)^\c \op\equiv\hints{by $\Ref{1}$ below ror the RHS}\hints{-- we choose the RHS because we don't seem to have}\hint{enough information about the LHS} \tag{*} h((\cup V)^\c) \;\subseteq\; \cap \{ B^\c \mid B \in V\} \op\equiv\hints{set theory: basic property of $\;\ldots \subseteq \cap \ldots\;$}\hint{-- the only way to make progress} \langle \forall B \in V :: \; h((\cup V)^\c) \subseteq B^\c \; \rangle \op\equiv\hint{$V \subseteq \T$ so $B \in \T$ or by $\Ref{0}$ equivalently $h(B^\c)=B^\c$} \langle \forall B \in V :: \; h((\cup V)^\c) \subseteq h(B^\c) \; \rangle \op\when\hint{by $\Ref{2}$ below -- this looks reasonable to try} \langle \forall B \in V :: \; (\cup V)^\c \subseteq B^\c \; \rangle \op\equiv\hint{set theory: basic property of $\;\ldots \subseteq \cap \ldots\;$} (\cup V)^\c \;\subseteq\; \cap \{ B^\c \mid B \in V\} \op\equiv\hint{by $\Ref{1}$ below} \true \endcalc$$

Here we used twice that $$ \tag{1} (\cup V)^\c \;=\; \cap \{ B^\c \mid B \in V\} $$ from set theory.

Now we are left to prove $$ \tag{2} A \subseteq B \;\then\; h(A) \subseteq h(B) $$ for any $\;A,B\;$. For this we calculate as follows, starting at the most complex side: $$\calc h(A) \subseteq h(B) \op\equiv\hints{set theory -- prepare for OP's property 2, which seems}\hint{the only information about $\;h\;$ we can use} h(A) \cup h(B) = h(B) \op\equiv\hint{OP's property 2} h(A \cup B) = h(B) \op\when\hint{logic -- just about the only thing we can do} A \cup B = B \op\equiv\hint{set theory} A \subseteq B \endcalc$$

And that completes the proof.

We see that this proof indeed it uses only OP's properties 2 and 3, and not property 4.

Comparing the structure of this write-up with the earlier one, we see that Mathematician 42's comment corresponds to the first calculation up until $\Ref{*}$; and the rest is the second part of their answer followed by its first part. Here, $\Ref{*}$ --expanded using $\Ref{1}$-- says that $\;h\;$ (i.e., taking the closure of a set) distributes over arbitrary intersections of complements of open sets (i.e., intersections of closed sets).

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Prove that $\mathcal T$ is a topology

3 Answers3