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This question already has an answer here:

If $A$ is a connected set and $\{A_i : i \in I\}$, $I$ an arbitrary set (can be countable or not) of connected sets.

How to show that if $A \cap A_i \neq \emptyset$ for all $i \in I$ then $A \cup (\cup_{i\in I} A_i)$ is connected?

I am trying to show that if $A \cap A_u \neq \emptyset~~ \forall i \in I$ then for all $i, j \in I$ $A_i\cap A_j \neq \emptyset.$ This enought to conclude the result. But I stuck here.

What I tried: Suppose that there are $i,j\in I$ such that $A_i \cap A_j = \emptyset.$

Then since $A\cap A_i \neq \emptyset$ and $A\cap A_j \neq \emptyset$ it somehow induces me to think that is possible to obtain a split for $A$. I don't know how to proceed.

Thanks

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marked as duplicate by Alex Provost, user99914, Henrik, Shailesh, Chill2Macht Jul 27 '16 at 1:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Not a duplicate, but sometimes called the patching lemma. Maybe there is a question related to that. $\endgroup$ – sqtrat Jul 26 '16 at 17:24
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That $A_i \cap A_j \not= \emptyset$ is not always true. Consider for each $n \in \mathbb{Z}$: $A_n:=\{n\}\times \Bbb{R}$ as a subspace of $\Bbb{R}^2$ and let $A=\mathbb{R}\times \{0\}$. Then $A_n\cap A \not= \emptyset$ for each $n$, but $A_n \cap A_m = \emptyset$ whenever $m\not=n$.

Let $X$ be a topological space. Then, show that every continuous function $f:X\rightarrow \{0,1\}$, where $\{0,1\}$ is endowed with the discrete topology, is constant if and only if $X$ is connected.

Then, consider a family $(A_i)_I$ of connected sets in $Y$ and $A$ connected such that $A_i \cap A \not= \emptyset$ for each $i \in I$.

Now, $j$ be an element not in $I$ and let $J=I \cup\{j\}$ and $A_j=A$. Then, let $X= \bigcup_J A_j$ and $f:X\rightarrow \{0,1\}$ be continuous. It's sufficient to prove that $f$ is constant. Let $x_1$ and $x_2$ be elements of $X$. Then, there is an $i_1,i_2$ in $J$ such that $x_k\in A_{i_k}$ for $k=1,2$.

Since there is an $y_k \in A_{i_k}\cap A$ for $k=1,2$, we have:

$$f(x_i)=f(y_i)$$ since $x_i,y_i \in A_{i_k}$ and $A_{i_k}$ connected. Then, since $A$ is connected we also have $f(y_1)=f(y_2).$

Thus $f(x_1)=f(y_1)=f(y_2)=f(x_2)$ and since $x_1$ and $x_2$ were arbitrary, we are done.

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  • $\begingroup$ I think that with questions like this (most probably exercises from a book/class) it's a shame to post a full solution---it takes away much of the learning opportunity for the OP. Perhaps consider phrasing your answer in the form of a hint? $\endgroup$ – Danu Jul 26 '16 at 17:18
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    $\begingroup$ I agree, but this is not the standard way of showing this result and is perhaps a little trickier to solve for the beginner. I therefore decided to prove the fact that it's constant and leave the part on the connectedness vs constant continuous functions as something the OP must show. $\endgroup$ – sqtrat Jul 26 '16 at 17:21
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Hint that should make it clear how to proceed: The connected component that contains $A$, in $A\cup\big(\bigcup_{i\in I} A_i\big)$, contains a point in each $A_i$ by your non-empty intersection condition.

If some connected component contains a point in $A_i$, then can it happen that it does not contain all of $A_i$?

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