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I have been working on the following problems, the first $3$ are from Dugundji's book, page $117$. I was wondering if someone can please check them or suggest easier proofs.

1) Let $\{A_{i}: i \in I\}$ be a family of connected subsets of $X$ and assume that there exists a connected set $A$ with $A \cap A_{i} \neq \emptyset$ for each $A_{i}$. Show $A \cup \bigcup_{i \in I} A_{i}$ is connected.

2) Let $\{A_{i}: i \in I\}$ be any family of connected sets. Assume that any two of them have non-empty intersection. Prove $\bigcup_{i \in I} A_{i}$ is connected.

3) Let $Y$ be a space and let $A \subset Y$ be any subset. Let $C \subset Y$ be connected such that $C$ intersects $A$ and $Y \setminus A$. Then it can be shown that $C$ contain points of $\textrm{bd}(A)$, the boundary of $A$.

Question: In $\mathbb{R}^{3}$, why is $A=\{(x,y,0) : x^{2}+y^{2} \leq 1\}$ and $C=\{(0,0,z): |z| \leq 1\}$ not a counterexample?

4) A doubt: $\mathbb{Q}$ is not locally connected.

My work:

1) For each $i \in I$ let $C_{i} = A \cup A_{i}$ then by assumption $C_{i}$ is connected. Now note that $A \subset \bigcap_{i \in I} C_{i}$. So $A \neq \emptyset$ for otherwise $A \cap A_{i} = \emptyset$, which contradicts the assumption. Thus $\bigcap_{i \in I} C_{i} \neq \emptyset$. Therefore $\bigcup_{i \in I} C_{i}$ is connected. But $\bigcup_{i \in I} C_{i} = A \cup \bigcup_{i \in I} A_{i}$.

2) Let $f: \bigcup_{i \in I} A_{i} \rightarrow \{0,1\}$ be a cts map. Now let $\alpha_{0} \in I$ be fixed and pick $\beta \in I$ such that $\beta \neq \alpha_{0}$. Then the restriction of $f$ to $A_{\beta}$ is constant (since $A_{\beta}$ is connected). So let $z_{\beta}=f(A_{\beta})$. By assumption $A_{\alpha_{0}} \cap A_{\beta}$ is non-empty so let $x$ be in the intersection. Then $f(x) \in f(A_{\alpha_{0}})$ and $f(x) in A_{\beta}$. Therefore $f(x)=z_{\beta}$ so $f$ is constant.

3) Is it because $C$ does not intersects the complement of $A$ or why?

4) Can we proceed as follows? suppose $\mathbb{Q}$ is locally connected and consider $(a,b) \cap \mathbb{Q}$ this is open in $\mathbb{Q}$. Now assuming $\mathbb{Q}$ is locally connected this would imply each connected component of $U$ is open. But $\mathbb{Q}$ is totally disconnected and $\mathbb{Q}$ is not discrete, so this is impossible.

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In your counterexample for question 3, $A$ has empty interior so $\mathsf{bd}(A)=A$. And if you were to just look at a disk with a line through the center, then the point where the disk intersected the line would also be a boundary point. –  JSchlather Jul 6 '11 at 19:56
@Jacob Schlather: Can you please explain why it has empty interior? –  user10 Jul 6 '11 at 20:13
Sure in $\mathbb R^3$ an open set must contain an open ball around each of it's points. But if you move $\epsilon$ away from any point on the disk in the $z$ direction, then you're not on the disk anymore. Another way is to note that the complement is dense in $\mathbb R^3$, so it must have empty interior. –  JSchlather Jul 6 '11 at 20:16
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3 Answers

up vote 1 down vote accepted

(1) is basically correct but a little clumsy. First, it isn't by assumption that $C_i$ is connected: it's by virtue of Theorem V.1.5 and the fact that $A \cap A_i \ne \emptyset$. Secondly, 'So' is inappropriate in your third sentence: it implies that what follows -- $A \ne \emptyset$ -- is a consequence of what you'd previously said, which isn't the case. Better: Clearly $A \ne \emptyset$, since $A$ has non-empty intersection with each $A_i$, so $\bigcap_{i \in I} C_i \ne \emptyset$.

(2) Basically fine. The wording could be improved a bit, and there's one bit of notational sloppiness that I really don't like: $f(A_\beta)$ is a set, so it equals $\{z_\beta\}$, not $z_\beta$.

(3) Use Dugundji's hint that $Y = \operatorname{Int}(A) \cup \operatorname{Fr}(A) \cup \operatorname{Int}(Y \setminus A)$ to conclude that if $C \cap \operatorname{Fr}A = \emptyset$, then $C \subseteq \operatorname{Int}(A) \cup \operatorname{Int}(Y \setminus A)$. In this case $C$ must intersect both $\operatorname{Int}(A)$ and $\operatorname{Int}(Y \setminus A)$; why, and why does this give you a 'disconnection' of $C$? (Jacob has already answered the other part of the question in his comment.)

(4) is fine; you're using Theorem V.4.2 and Ex. 1 on p. 111.

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1,2 and 4 look fine, assuming that for 4 you can assume $\mathbb Q$ is totally disconnected (this isn't hard to prove regardless). A nicer way of looking at that question (and indeed what you did) is that a locally connected space is totally disconnected if and only if it is has the discrete topology.

For question 3 I think the easiest way to go about it is to suppose that $\mathsf{bd}(A) \cap C$ is empty. Then what can you say about $\mathsf{Int}(A)$ and $Y \setminus \overline{A}$ with regards to $C$?

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Questions (1)--(3) can all be handled by appealing to the topological definition of "connected"... That is, if I want to prove $X$ is connected, then I assume $X \subset U \cup V$ for open sets $U$, $V$ such that $U \cap V = \emptyset$. If I can logically show that $X$ must be contained only in $U$ or only in $V$, then I have proven $X$ connected. This type of proof often starts by observing some given connected set $C$ is in $U \cup V$, and therefore we know that $C$ lies only in $U$ (we don't have to consider $C$ lies only in $V$. Why not?).

For your question (3), the example you have given is not complete. I don't know what $Y$ is. Whatever $Y$ is, it must properly contain $A$ (so that $Y \setminus A$ is non-empty), so let's just assume $Y = \mathbb{R}^3$ for the moment. Then the hypotheses of (3) are satisfied. The conclusion is also satisfied, since $C$ contains the point $(0,0,0)$. This is a boundary point of $A$, since $A \subset \mathbb{R}^3$! (It would not be a boundary point if we only worked in $\mathbb{R}^2$).

For (4), I'd argue from definitions again. We say that $X$ is locally connected at $x$ if for every open set $V$ containing $x$ there exists a connected, open set $U$ with $x \in U \subseteq V$. $\mathbb{Q}$ is not locally connected at $0$ since for $V = (-1,1)$, any open set $U$ containing $0$ such that $U \subseteq V$ consists of unions of open intervals within $(-1,1)$. Just consider the open interval $(a,b) \subseteq U$ containing $0$. There must be a negative rational and a positive rational. Can you find an irrational number lying between these two? Such an irrational number would be a "hole" that can be used to define a separation of the interval $(a,b)$.

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@Jacob Schlather But totally disconnected is not equivalent to discrete, as the example $\mathbb{Q}$ shows. If $\mathbb{Q}$ were discrete, then singletons $\{a\}$ would be open sets! -------OOPS, just re-read the comment and saw "locally connected" before that if-and-only-if. You're correct :) –  Shaun Ault Jul 6 '11 at 20:28
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