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I have to give necessary and sufficient conditions s.t. $ X=\prod_i X_i$ is locally / strongly connected, where $(X_i,\tau_i)$ are non-empty top. spaces.

First locally: Assume all $X_i$ are locally connected. Let $x \in X$. Consider some fixed $i$. Then $x_i = p_i(x) \in X_i$. So there is some connected $x \in U \in \tau_i$. Then $x \in p_i^{-1}(U)$ where $p_i^{-1}(U)$ is connected and open since $p_i,p_i^{-1}$ are continuous ($p_i$ is open). So $X$ is locally connected.

On the other hand, if $X$ is locally connected: Consider some arbitrary $i$. Let $x_i \in X_i$. Then there is some $z \in X$ with $z_i = x_i$, since all $X_k$ are non-empty. Now there is a connected $z \in U \in \tau$ (product topology) and $p_i(U)$ is a connected neighborhood of $x_i$ in $X_i$ since $p_i,p_i^{-1}$ are continuous.

Now strongly: Assume $X$ is strongly connected, then if $x_i \in X_i$ let $U$ be an open neighborhood of $x_i$. Let $z$ again in $X$ with $z_i = x_i$. Then $z \in p_i^{-1}(U) \in \tau$. So there is some connected $V \in \tau$ with $z \in V \subset p_i^{-1}(V)$. So $x_i \in p_i(V) \subset U$ where $p_i(V)$ is connected and open in $X_i$.

The other way round: Given $x \in U \in \tau$ we can find $i_1,\cdots,i_n$ and $U_1 \in \tau_{i_1},\cdots,U_n \in \tau_{i_n}$ s.t. $$ x\in \prod_{k=1}^n U_{k} \subset U. $$ Let $1 \leq k \leq n$. Then $x_{i_k} \in U_{k}$. Assuming all $X_i$ are strongly connected we see that there is some $U_k' \subset U_k$ which is conneted and open. Now $$ x\in \prod_{k=1}^n U_k' \subset \prod_{k=1}^n U_k \subset U $$ and the first product is connected and open since it is a finite product.

So in both cases I guess $X$ locally / strongly conn. iff $\forall i: X_i$ locally / strongly conneted. Didn't I make any mistakes ?

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What are your definitions of local and strong connectedness? – Luiz Cordeiro Jan 15 '14 at 11:37
Locally if the property holds for some open neighborhood, strongly if there is a neighborhoodbasis of such elements. It should be weakly locally and strongly locally I recognize now. – Epsilon Jan 15 '14 at 12:19
The proof of the local connectedness of the product is not correct. You consider only a finite product, but you still have to multiply with $X_j$ for all $j\ne i_1,...,i_n$. It would work if all spaces were also connected, though. – Stefan Hamcke Mar 3 '14 at 0:04
For example, $\{0,2\}$ with the discrete topology is locally connected, but the Cantor Set $C=\{0,2\}^\omega$ is totally separated, thus also totally disconnected. As every point is a limit point, no point has a connected neighborhood. – Stefan Hamcke Mar 3 '14 at 15:46
@André Stefan is perfectly right. In fact, your product space is locally (weakly or strongly) connected if and only if all spaces $X_i$ are locally connected and all but finitely many $X_i$ are connected. – Etienne Mar 9 '14 at 14:17

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