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I was working with this Theorem and somehow I don't see clearly how it's being proved.

Theorem: Let $(X, \tau)$ be topological space and $A_\alpha \subset X$ be connected for every $\alpha \in I$, such that $ X=\displaystyle \cup_{\alpha \in I} A_\alpha$. If there exists a $\alpha_0 \in I$ such that for every $\alpha \in I$ , $A_{\alpha_0} \cap A_\alpha \neq \emptyset $, then $X$ is connected.

Proof: Suppose $X$ is not connected, we will see that if $A_\alpha$ is connected for every $\alpha \neq \alpha_0$, then $A_{\alpha_0}$ is not connected.

Let $(A,B)$ open sets such that they are disjoint, nonempty and their union is $X$, we'll prove that $A \cap A_{\alpha_0} \neq \emptyset , B \cap A_{\alpha_0} \neq \emptyset$ ( meaning $A \cap A_{\alpha_0}$ and $ B \cap A_{\alpha_0} $ also satisfy the previous requirements * ). Wlog, suppose $A \cap A_{\alpha_0} \neq \emptyset$ because $A_\alpha$ is connected for every $\alpha \neq \alpha_0$, and by prop(2), either $A_\alpha \subset A$ or $A_\alpha \subset B$. Let $\eta=\{ \alpha \in I: A_\alpha \subset B\}$, note that $\eta \neq \emptyset$ otherwise $A_\alpha \subset A, \forall \alpha \in I$ and $B=B \cap (\displaystyle \cup_{\alpha \in I} )$ would be empty. Take $\alpha \in \eta$ this implies $A_\alpha \subset B$ and therefore $\emptyset \neq A_\alpha \subset B \subset A_\alpha \cup A_{\alpha_0} \subset B \cap A_{\alpha_0} $

(*) Also means that they're proving $A_{\alpha_0}$ is a disconnected subset of $X$.

Proposition 2 Let $A, B$ open sets, disjoint, nonempty and $X = A \cup B$ for $X$ a topological space, and let $C \subset X$ a connected subset. Then $C \subset A$ or $C \subset B$.

I know is very straightforward with respect to prove that $A_{\alpha_0}$ it's a disconnected subset, I get it, but how do you come with the idea that proving $X$ is connected means to prove something is a disconnected subset? More precisely the bold part.

Thank you in advance.

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  • $\begingroup$ take a look here. In short: prove that $A\implies B$ is equivalent to prove that $\lnot B\implies\lnot A$. $\endgroup$ – Masacroso May 28 '18 at 5:44
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The part in bold is a variation of the contrapositive of the theorem.

The theorem states that if $A_\alpha$ is connected for every $\alpha \in I$ and there exists $\alpha_0 \in I$ such that, for every $\alpha \in I$, $A_{\alpha_0} \cap A_\alpha \neq \emptyset$, then $X=\bigcup_{\alpha \in I}A_\alpha$ is connected.

The contrapositive of this statement is, if $X$ is not connected, then either there exists an $\alpha \in I$ such that $A_\alpha$ is not connected or for every $\alpha_0 \in I$, there exists $\alpha \in I$ such that $A_{\alpha_0} \cap A_\alpha = \emptyset$. This statement is equivalent to the theorem.

Note the since $P$ or $Q$ is equivalent to (not $Q$) implies $P$, we can rewrite the contrapositive like so.

If $X$ is not connected and there exists $a_0 \in I$ such that $A_{\alpha_0} \cap A_\alpha \neq \emptyset$, then there exists $\alpha \in I$ such that $A_\alpha$ is not connected. Again this statement is equivalent to the theorem.

Finally the statement in bold is equivalent to If $X$ is not connected and there exists $\alpha_0 \in I$ such that $A_{\alpha_0} \cap A_\alpha \neq \emptyset$, then either $A_\alpha$ is not connected for some $\alpha \neq \alpha_0$ or $A_{\alpha_0}$ is not connected. This is again equivalent to the theorem since for every $\alpha \in I$ either $\alpha=\alpha_0$ or $\alpha \neq \alpha_0$.

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  • $\begingroup$ Thanks, where can I see that $P or Q \leftrightarrow not P \rightarrow$ I mean such equivalence. Hehe $\endgroup$ – Lilian Hernández May 28 '18 at 13:48
  • $\begingroup$ This question talks about $P \rightarrow Q \leftrightarrow \neg P \lor Q$ math.stackexchange.com/q/935640. If we subsitute $\neg P$ for $P$, we get the result. $\endgroup$ – michaelhowes May 28 '18 at 21:07
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This is reductio ad absurdum, or proof by contradiction...

The conclusion of the theorem, namely that $X$ is connected, is assumed to be false.

Next a contradiction to the assumptions of the theorem is sought.

Once this contradiction, namely that $A_{\alpha_0}$ is not connected, is obtained, we know our assumption that $X$ is not connected must have been false...

Hence the contrary is true, or $X$ is connected.

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  • $\begingroup$ Am, do you mean that if I said that every $A_\alpha$ was connected and because their union was $X$, if there exists one which is not connected represents a contradiction? A set is always either formed of connected sets or not connected, but not both? $\endgroup$ – Lilian Hernández May 28 '18 at 7:07
  • $\begingroup$ @LilianHernández The contradiction is that one of the $A_{\alpha}$, namely $A_{\alpha_0}$, is found to not be connected... they were all assumed to be connected... $\endgroup$ – Chris Custer May 28 '18 at 7:21
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To show a topological space $X$ is connected, you have to show that for any open set $U$ of $X$, if the complement of $U$ is also open then $U = X$ or $U = \emptyset$. Seems like a whole bunch of work, checking something for every open subset! So when proving a general proposition you want to get your hands on something more concrete, like assuming that a space is not connected and equipped with a nontrivial decomposition. So parsing out the contrapositive of the OP's proposition is the way to go.

This can get tricky when working with compound math statements that involve quantifiers. I am not completely happy with the proof the OP gives; the statement

Wlog, suppose $A \cap A_{\alpha_0} \neq \emptyset$ because $A_\alpha$ is connected for every $\alpha \neq \alpha_0$, and by prop(2), either $A_\alpha \subset A$ or $A_\alpha \subset B$.

leaves me asking

$\quad \text{What if } I = \{\alpha_0\}?$

So allow me to sketch out my own 'contrapositive parse', with two math sections.


Section 1

For the remainder of this section, $X$ is a set partitioned into two subsets $U$ and $V$.

Lemma 1: Let $A_1$ and $B_2$ be two subsets of $X$ with $A_1 \subset U$ and $A_2 \subset V$. If $A_0$ is any subset of $X$ that has nonempty intersections with both $A_1$ and $A_2$, then the following also holds:

$\tag 1 A_0 \cap U \ne \emptyset \text{ and } A_0 \cap V \ne \emptyset$

Lemma 2: Let $A_0$ and $A_1$ be two nonempty subsets of $X$ satisfying the following:

$\tag 2 X = A_0 \cup A_1 \text{ AND } A_1 \subset U \text{ or } A_1 \subset V \text{ AND } A_0 \cap A_1 \ne \emptyset$

Then $\text{(1)}$ holds true (again).

Lemma 3: Let $(A_\alpha)$ be a family of nonempty subsets of $X$ where the index set $I$ for $\alpha$ contains at least two distinct elements, $\alpha_0$ and $\alpha_1$, satisfying the following,

$\tag 3 X = \bigcup_{\alpha} A_\alpha$

$\tag 4 \text{For every } \alpha \in I \text{, IF } \alpha \ne \alpha_0 \text{ THEN } A_\alpha \subset U \text{ or } A_\alpha \subset V$

$\tag 5 \text{For every } \alpha \in I \text{, IF } \alpha \ne \alpha_0 \text{ THEN } A_{\alpha_0} \cap A_{\alpha} \ne \emptyset$

Then, setting $A_0 = A_{\alpha_0}$, $\text{(1)}$ holds true (now for the third time).

Proof(sketch)

If you can't wrap things up with lemma 1, then take $A_0 = A_{\alpha_0}$ and let $A_1 = \cup A_\alpha$ with $\alpha \ne \alpha_0$ and apply lemma 2.


Section 2

Theorem: Let $(X, \tau)$ be topological space and $A_\alpha \subset X$ be connected for every $\alpha \in I$, such that $ X=\displaystyle \cup_{\alpha \in I} A_\alpha$. If there exists a $\alpha_0 \in I$ such that for every $\alpha \in I$ , $A_{\alpha_0} \cap A_\alpha \neq \emptyset $, then $X$ is connected.

Proof

If the index set $I$ is empty or a singleton set equal to $\{\alpha_0\}$, then the theorem is trivial. So we assume that the $I$ has a least one other element besides $\alpha_0$.

To prove the contrapositive of the theorem, we can apply lemma 3 from the set-theoretic machinery developed in section 1.

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