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Definition. Let $(X,\tau)$ be a topological space and $a,b\in X$. We will say that $a$ and $b$ are connected by a simple chain of open sets if there exists open sets $U_1,U_2,\ldots,U_n$ such that $a\in U_1$ only and $b\in U_n$ only and $U_i\cap U_{j}\ne\emptyset$ iff $|i-j|\le 1$.

Then we know that $(X,\tau)$ is connected iff given any open cover $\mathscr{U}$ of $X$ and any two points $a,b\in X$, there exists a simple chain of open sets of $\mathscr{U}$ connecting $a$ and $b$. (See this for clarification.)

Let $(X,\tau)$ be a topological space and $\mathscr{U}$ be an open cover of $X$. define a relation $\sim_{\mathscr{U}}$ on $X$ as follows, $$a\sim_{\mathscr{U}} b\iff a\ \text{and}\ b\ \text{are connected via a simple chain of open sets from}\ \mathscr{U}$$

Question

Is $\sim_{\mathscr{U}}$ a transitive relation on $X$?

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2 Answers 2

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Hint: being connected by a chain of open sets (the same thing, but with condition restricted to $\lvert i-j\rvert\leq 1\implies U_i\cap U_j\neq\emptyset$) is clearly transitive. Now note that if $U_1,\ldots,U_n,\ldots,U_k,\ldots,U_m$ is a chain and $U_n\cap U_k\neq \emptyset$, then $U_1,\ldots,U_n,U_k,\ldots,U_m$ is also a chain.

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  • $\begingroup$ I am sorry. But I don't understand your hint. More specifically, let $a\sim_{\mathscr{U}} b$ and $b\sim_{\mathscr{U}} c$ and let $\{U_1,\ldots,U_n\} $and $\{V_1,\ldots,V_m\}$ be the respective simple chains. Then what's your point exactly? $\endgroup$ Commented Mar 9, 2018 at 15:40
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    $\begingroup$ Concatenate the chains into $(U_1,\dots,U_n,V_1,\dots,V_m)$ and simplify it until it is simple. $\endgroup$ Commented Mar 9, 2018 at 15:43
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You write

"Then we know that $(X,\tau)$ is connected iff given any open cover $U$ of $X$ and any two points $a,b \in X$, there exists a simple chain of open sets of $U$ connecting $a$ and $b$."

Do you really know that? Let $X =\{a, b\}$, and $U = \{ \emptyset, X, \{b\}\}.$

It looks to me as if this $X$ is connected, but there's no chain from $a$ to $b$ because you've required that $a \in U_1$ only, and $b \in U_2$ only.

Maybe I don't understand what "only" modifies here. Does it mean that among all the $U_i$, the only one that contains $a$ is $U_1$? Or does it mean that $U_1$ contains $a$ but not $b$, while $U_n$ contains $b$ but not $a$.

In either case, this example seems to fail. So it's tough to answer the rest of your question. But for any reasonable interpretation I can put on it, it appears to me that in the topology I described, it's not true that $a \sim_U a$.

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  • $\begingroup$ See this for clarification. I have taken the result from this book. $\endgroup$ Commented Mar 9, 2018 at 15:23
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    $\begingroup$ I think "only" means that, among all the $U_i$, the only one that contains $a$ is $U_1$, and the only one that contains $b$ is $U_n$. In particular, let our chain have a single element, $U_1:=\{a,b\}$. Then $n=1$ and $U_n=U_1$, and $a\in U_1$ only and $b\in U_n$ only. So they are connected by a simple chain. $\endgroup$ Commented Mar 9, 2018 at 15:33
  • $\begingroup$ Ah ... a third possible meaning for "only" in this context, and one that makes sense. Thanks! I think I'll leave my answer so that your interpretation remains. :) $\endgroup$ Commented Mar 9, 2018 at 17:11
  • $\begingroup$ That 'only' doesn't alter the result actually because if you take any simple chain that connect $a$ and $b$, then $a$ can be in $U_1$ and $U_2$ only and in that case we can start our chain from $U_2$. Similarly, for $b$. $\endgroup$ Commented Feb 13, 2022 at 17:32

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