I'm teaching my self topology using a book I found. This is the forth part of a 4 part question. links to other parts: one, two, three .

I'm trying to prove the following problem from a book I found:

Let $X$ be a topological space and let $\mathscr{A}$ be a collection of subset of $X$. Prove
$int( \bigcup \limits_{A \in \mathscr{A}} A)\supseteq \bigcup \limits_{A \in \mathscr{A}} int(A)$

I want to know if my proof is valid, and before i prove it I will given an example of $int( \bigcup \limits_{A \in \mathscr{A}} A)\supset \bigcup \limits_{A \in \mathscr{A}} int(A)$.

Example of $int( \bigcup \limits_{A \in \mathscr{A}} A)\supset \bigcup \limits_{A \in \mathscr{A}} int(A)$.

Let: $\mathscr A =\{[1,2] , [2,3]\}$. So then, $int( \bigcup \limits_{A \in \mathscr{A}} A) =(1,3)$ and $ \bigcup \limits_{A \in \mathscr{A}} int(A)=(1,3)\setminus{2}. $

Proof of: $int( \bigcup \limits_{A \in \mathscr{A}} A)\supseteq \bigcup \limits_{A \in \mathscr{A}} int(A)$

To prove this, I will use 3 corollaries, that will not be proven here.
(1) $ \forall S \in \mathscr S : S \subseteq \bigcup \limits_{S \in \mathscr{S}} S$; Any individual set in a collection is a subset of the union of the entire collection.
(2) $U \subseteq V \implies int(U) \subseteq int(V)$ : IF U is a subset of V, THEN the interior of U is a subset of the interior of V.
(3) $\forall X \in \mathscr X : X \subseteq Y \implies \bigcup \limits_{X \in \mathscr{X}} X \subseteq Y$: IF If every subset in a collection is a subset of Y THEN the union of the collection is also a subset of Y.

Proof: Let, $\mathscr A$ be any collection of sets in a topological space, and let $A$ be any individual set in the collection.

$ \forall A \in \mathscr A : A \subseteq \bigcup \limits_{A \in \mathscr{A}} A$; (1, $S=A$)

$ \forall A \in \mathscr A : int(A) \subseteq int(\bigcup \limits_{A \in \mathscr A} A )$; (2, $U = A $ and $V= \bigcup \limits_{A \in \mathscr A} A $ and )

$\bigcup \limits_{A \in \mathscr{A}} int(A) \subseteq int( \bigcup \limits_{A \in \mathscr{A}} A)$ ; (3, $X=int(A)$ and $Y=int(\bigcup \limits_{A \in \mathscr A} A )$)


Is this proof valid? Am I using the $\forall$ symbol correctly?

  • $\begingroup$ Your proof is valid, though you should better write $$∀A∈A:A⊆B$$ or $$A⊆B \text{ for all }A∈A$$ instead of $$\color{red}{\forall A\subseteq B}$$ $\endgroup$ – Stefan Hamcke Jul 13 '16 at 20:31
  • $\begingroup$ The same proof applies to the closure operator, or any operator with the property (2). Such an operator is said to be monotone (or monotonic, or order-preserving). $\endgroup$ – Stefan Hamcke Jul 13 '16 at 20:36
  • $\begingroup$ Parts 1 and 2 are for the closure. I proved those is a less elegant manner. I used the answer by @user254665 for part 2, to do 3 and 4. I spent over 2 weeks thinking about this problem, and mainly thanks to user254665 I feel like a finally have a grasp on this concept. $\endgroup$ – Michael Maliszesky Jul 13 '16 at 20:41

Answers to questions: Yes and No, in that order. For the second question, your mistake is that you are overlapping the quantifier scope with the formula. Example: you write the first step of the proof (with justification removed) as:

$ \forall A \subseteq \bigcup \limits_{A \in \mathscr{A}} A$

What you should be doing is quantifying the variable separately first, then giving the condition satisfied by the variable, i.e:

$ \forall A \in \mathscr{A}$ : $ A \subseteq \bigcup \limits_{A \in \mathscr{A}} A$

Similarly, the second step should be written:

$\forall A \in \mathscr{A}$ : $ int(A) \subseteq int(\bigcup \limits_{A \in \mathscr A} A )$

and so forth.

  • $\begingroup$ thanks, i just made the edits. I think its good now. $\endgroup$ – Michael Maliszesky Jul 13 '16 at 21:05

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