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Is it true that a topological space $X$ is connected if and only if for every open cover $\{U_s\}_{s \in S}$ of $X$, and for every pair of points $x_1,x_2 \in X$, there is a finite sequence $s_1,...,s_k$ of elements of $S$ such that $x_1 \in U_{s_1} , x_2 \in U_{s_k}$ and $U_{s_i} \cap U_{s_j} \ne \emptyset$ iff $|i-j|\le1$ ?

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HINT: Let $X$ be a topological space. For $x_1,x_2\in X$ write $x_1\sim x_2$ if and only if there is a finite sequence as in the statement of the question.

  • Show that $\sim$ is an equivalence relation on $X$.
  • Show that each equivalence class of $\sim$ is a union of open sets and is therefore open.
  • Explain how the desired result follows from this.
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  • $\begingroup$ I recently solved this exercise (or so I thought). Define, por each $p\in X$ the set $\{q\in X\mid \text{ there is a finite sequence as stated }\}$. This set is non-empty, open and closed, so, because $X$ is connected, then this set must be equal to $X$. For the converse implication assume $X$ is disconnected, then there are open sets $U$ and $V$ such that $U\cup V=X$. Tales the cover $\mathcal{U}=\{U,V\}$. This cover doesn't satisfy the statement give, so not connected implies not statement. $\endgroup$ – user419934 May 27 '17 at 22:32
  • $\begingroup$ But then I ran into this problem, imagine $X=U\cup V\cup W$, $U$, $V$, and $W$ open. What happens if $V$ is disconnected and $U\cap V\neq \emptyset$, $V\cap W\neq \emptyset$. This cover satisfies the statement given but $Y$ would be disconnected. What am I not seeing? $\endgroup$ – user419934 May 27 '17 at 22:33

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