$T_1$ space $X$ is both connected and locally connected iff for every open cover $\{U_\alpha\}$ of $X$ and pair of points $x_1,x_2$ of $X$, there exists a finite sequence $\alpha_1,\cdots,\alpha_n$ and a sequence of connected open subsets $V_1,\cdots,V_n$ such that
- $x_1\in V_1$, $x_2\in V_n$
- $V_i\cap V_j\neq\varnothing$ iff $|i-j|\leq1$
- $V_i\subseteq U_{\alpha_i}$ for all $i=1,\cdots,n$
Now, for connected $X$, we have that for $x_1,x_2$ of $X$ and open cover $\{U_\alpha\}$, we can get a sequence $U_{\alpha_1},\cdots,U_{\alpha_n}$ from the cover such that
- $x_1\in U_{\alpha_1}$, $x_2\in U_{\alpha_n}$
- $U_{\alpha_i}\cap U_{\alpha_j}\neq\varnothing$ iff $|i-j|\leq1$
Also, as $X$ is locally connected, each component of an open set is open.
Now, I believe that the $V_i$ required are components of $U_{\alpha_i}$, appropriately chosen so that Conditions $1$ and $2$ hold. This would automatically take care if condition $3$. However, I haven't been able to show this. Any help would be appreciated!
Best Answer
Let $X$ satisfy the "open cover condition". Then $X$ is connected because any two $x_1, x_2 \in X$ are contained in a connected subset of $X$ (take the union of the $V_i$). To show that $X$ is locally connected, let $x_1 \in X$ and $U_1$ be an open neighborhood of $x_1$. We have to find a connected open neighborhood $V_1$ of $x_1$ such that $V_1 \subset U_1$. The set $U = X \setminus \{x_1\}$ is open since $X$ is $T_1$ (this is the only place where we need the $T_1$-requirement). Therefore $\mathcal U = \{U_1, U\}$ is an open cover of $X$. Choose any $x_2 \in X$ (if you want $x_2 = x_1$). There exists a sequence of connected open $V_i$ as in your condition. We have $x_1 \in V_1$. Moreover, $V_1$ is contained in some member of $\mathcal U$. Since $x_1 \in V_1$, it impossible that $V_1 \subset U$. Thus $V_1 \subset U_1$.
We next prove the converse. Let us begin with the following
Lemma: Let $M_1,\ldots, M_r$ be subsets of $X$ such that $M_i \cap M_{i+1} \ne \emptyset$ for $i =1,\ldots,r-1$. Then there exists a subset $\{k_1,\ldots,k_n\} \subset \{1,\ldots,r\}$ such that $1 = k_1 < k_2 <\ldots k_{n-1} < k_n=r$ and $M_{k_i} \cap M_{k_j} \ne \emptyset$ iff $\lvert i - j \rvert \le 1$.
Proof: Call $\{k_1,\ldots,k_n\} \subset \{1,\ldots,r\}$ nice if $1 = k_1 < k_2 <\ldots k_{n-1} < k_n=r$ and $M_{k_i} \cap M_{k_{i+1}} \ne \emptyset$ for $i = 1,\ldots,n-1$. Clearly $\{1,\ldots,r\}$ is nice. There exists a nice $\{k_1,\ldots,k_n\}$ with minimal $n$ (possibly $n = r$). Assume $M_{k_i} \cap M_{k_j} \ne \emptyset$ for some pair $(i,j)$ such that $\lvert i - j \rvert > 1$. W.l.o.g. we may assume $i < j$. Then $\{k'_1 = k_1,\ldots,k'_i = k_i,k'_{i+1} = k_j,\ldots,k'_{n+1-(j-i)} = k_n\}$ is nice with $n+1-(j-i) < n$, a contradiction.
The lemma shows that in the "open cover condition" we can replace 2. by the (only apparently) weaker condition $$V_i \cap V_{i+1} \ne \emptyset, i =1,\ldots,n-1 .$$ Let $\mathcal U$ be an open cover of $X$. For $x_1,x_2 \in X$ define $x_1 \sim x_2$ if there exists a finite sequence of connected open subsets $V_1,\cdots,V_n$ such that
$\sim$ is an equivalence relation. Reflexivity is due to local connectedness (each $x$ is contained in some $U \in \mathcal U$, now take $n=1$ and $V_1$ any connected open such that $x \in V_1 \subset U$). Symmetry and transitivity are obvious.
The equivalence classes $[x_1]$ with respect to $\sim $ are open: If $x_2 \in [x_1]$, we find a sequence of $V_i$ as above. But then obviously $x_2 \in V_n \subset [x_1]$. Hence the equivalence classes form a partitioning of $X$ into pairwise disjoint open sets. Since $X$ is connected, there can be only one equivalence class. Thus any two $x_1,x_2 \in X$ are equivalent which finishes the proof.