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Let $X$ be a space. Let $\mathscr{B}$ be a covering of $X$ and let each element in $\mathscr{B}$ be a connected subset of $X$. Suppose that if A and B are in $\mathscr B$ then there is a finite subcollection {$B_1, B_2,..., B_n$} such that A=$B_1$ and B=$B_n$ and $B_i\cap B_{i+1}$$\not=$$\varnothing$ for i = 1,2,...,n-1. Prove that X is connected.

Here is what I have

Since $\mathscr B$ is a cover of $X$, the union of all $\mathscr B$ elements contains $X$. But, since each element in $\mathscr B$ is a subset of $X$ its union is contained in $X$. Thus the union of all $\mathscr B$ elements is equal to $X$.

(If $\mathscr B$ is finite I would be finished. But it is not) My idea from here is as such. Let A be the smallest non empty set in $\mathscr B$ and let B = $X$ Then there is a finite set from $\mathscr B$ that adheres to the last part of the hypothesis and whose union is equal to X. Thus $X$ is connected by this theorem: Let $\mathscr A$ be a collection of connected subspaces of a topological space ($X,\mathscr T$), and let $A$ equal the union of $\mathscr A$ elements. Then if $\mathscr A$ is a finite set and the intersection of any two sets from $\mathscr A$ is non empty, then $A$ is connected

Is this correct? If so, is it efficient? If not, any help would be appreciated

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  • $\begingroup$ By "finite subcollection" do you mean finite subcover or just a collection of sets? Also, did you mean to write $B_i\cap B_{i+1}\neq \emptyset$? $\endgroup$ – user160738 Feb 14 '17 at 1:03
  • $\begingroup$ Juat a collection of sets and yes I did sorry thank you $\endgroup$ – tike baylor Feb 14 '17 at 1:11
  • $\begingroup$ I also just realized that "Let A be the smallest..." Does not cover all possibilities for A and B $\endgroup$ – tike baylor Feb 14 '17 at 1:18
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Hint:

(1). Show that $\cup_{i=1}^nB_i$ is connected if each $B_i$ is connected and if $B_i\cap B_{i+1}\ne \phi$ for $1\leq i<n.$

(2). Suppose $X=Y\cup Z$ where $Y,Z$ are disjoint, open, and not empty. Let $a\in Y$ and $b\in Z$ and take $B_1,..., B_n$ as in (1) with $a\in B_1$ and $b\in B_n,$ and obtain a contradiction.

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  • $\begingroup$ Thank you. I think I see it now. $\endgroup$ – tike baylor Feb 14 '17 at 2:17
  • $\begingroup$ Did you mean a $\in B_1$ $\endgroup$ – tike baylor Feb 14 '17 at 13:25
  • $\begingroup$ Yes. Thanks. It was a typo. $\endgroup$ – DanielWainfleet Feb 14 '17 at 20:23

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