Prove the statements about connectedness are equivalent Definition:A space $X$ is connected if and only if $X$ is not the union of two disjoint non-empty open sets.
Prove $X$ is connected if and only if, for every pair of points $p$ and $q$ and every open cover $\{U_\alpha\}_{\alpha \in \lambda}$ of $X$ there exist a finite number of $U_{\alpha}$'s ,$\{U_{\alpha_1},...,U_{\alpha_n}\}$ such that $p \in U_{\alpha_1},q \in U_{\alpha_n}$ and for each $i<n$,$U_{\alpha_i} \cap U_{\alpha_{i+1}}\neq \varnothing$.
I was attempting this and trying to negate the second part of the statement in order to prove the contrapositive in both directions.The negation I came up with is the following:
There exist points $p$ and $q$ and an open over $\{U_\alpha\}$ of $X$ such that for all finite subcollections $\{U_{\alpha_1},...,U_{\alpha_n}\}$ such that $p \in U_{\alpha_1},q \in U_{\alpha_n}$, there is an $i <n$ with $U_{\alpha_i} \cap U_{\alpha_{i+1}}=\varnothing$.
Now I will try to attempt the proof by proving $X$ is not connected $\iff$ the statement directly above this sentence.
Suppose $X$ is not connected. Let $U,V$ constitute a pair of disjoint nonempty open sets whose union is $X$. Then $U \cup V=X$ so that $\{U,V\}$ is an open cover of $X$.
For the subcollection $\{U,V\}$ of $\{U,V\}$, choose any point $p \in U$ and $q \in V$. Set $U=U_{\alpha_1}$ and $V=U_{\alpha_2}$. Then for $i=1<2$, $U_{\alpha_1} \cap U_{\alpha_2}=\varnothing$.
For any other subcollection of $\{U,V\}$, the conditions of the conclusion of the statement are automatically satisfied.
Now suppose that the conditions in the fourth paragraph are satisfied. Let $p,q$ be such points and $\{U_\alpha\}$ be such an open cover. Need to show $X$ is not connected.
I am now having trouble with this part of the proof. Any suggestions? Also was my try at the first part of the proof correct?
 A: I already showed that fact in this answer, but again:
If the condition holds, then $X$ is connected: if $X$ were not, then $X= U \cup V$ where $U,V$ are non-empty, open and disjoint. But then for the open cover $\{U,V\}$ of $X$ we cannot have any "link-chain" (as such a finite family of open sets is called sometimes) from anu $p \in U$ to any $q \in V$, simply because there are no distinct intersecting open sets in this cover at all. This roughly agrees with your argument.
So for the interesting direction, assume $X$ is connected, $\mathcal{U}$ is an open cover and $p,q \in X$. We can proceed quite directly: define $$O = \{x \in X\mid \exists n \in \Bbb N, \exists U_1, U_2, \ldots U_n \in \mathcal{U}: p \in U, x \in U_n, \forall i < n: U_i \cap U_{i+1} \neq \emptyset\}$$
So we want to show $q \in O$. This follows as $O$ is open, closed and non-empty and so by connectedness of $X$ can only be $X$ (and $q \in X$):

*

*$O$ is non-empty as $p \in O$ as witnessed by the "chain" $\{U_1\}$, any set from $\mathcal{U}$ that contains $p$. The linking condition is void for a chain of length one.


*$O$ is open: pick any $x \in O$ and a witnessing chain $U_1, U_2, \ldots ,U_n \in \mathcal{U}$ for $x$ so that $p \in U_1, x \in U_n$ and all consecutive sets intersect. Then for any $y \in U_n$, the same chain witnesses that $y \in O$ as well, so $x \in U_n \subseteq O$ and $x$ is an interior point of $O$, and as $O$ is arbitrary, $O$ is open.


*$O$ is closed: let $y \in \overline{O}$. Let $U$ be any set in $\mathcal{U}$ that contains $y$ (using again that we have a cover of $X$), and as $U$ is an open neighbourhood of $y$, $U \cap O \neq \emptyset$, say $x \in U \cap O$. As $x \in O$ there is a witnessing chain $U_1, \ldots, U_n$ (intersecting etc. and $p \in U_1, x \in U_n$) and then $U_1, \ldots, U_n,U$ is a chain that witnesses $y \in O$ too (as the added intersection is non-empty by construction and $y \in U$). So $\overline{O} \subseteq O$ so $O$ is closed.
QED.
