Product of connected spaces You have two connected topological spaces $(A,B)$. Prove that $A\times B$ is also connected.
I understand that I have to prove that there is a point in $B$ (call it $b$), that makes $A\times\{b\}$ homeomorphic to $A$ making it connected to $A\times B$. Then prove that $\{a\}\times B$ is connected in $A\times B$. But I don't really know where to being with this. If you could help that would be appreciated.
 A: Here's another approach. I think it's interesting because like many phenomena in geometry it deals with "propagation" of properties from the base space "through" the fibers.
First prove the following.
Proposition. Let $X\overset{f}{\to} Y$ be an open set-function between topological spaces. Suppose $Y$ is connected and the nonempty fibers of $f$ are connected. Then $X$ is connected.
Next use the fact $A\times B\to B$ is open (as a product projection), and observe its nonempty fibers are homeomorphic to $A$, and therefore connected. Since $B$ is also connected (by assumption), the proposition implies the desired result.
A: Let $F : A \times B \to \{0,1\}$ be a continuous function. To show that $A\times B$ is connected for the product topology we have to show that $F$ is constant.
As you suggested (kind of) we first show that $F$ is constant on every set of the form $\{a\}\times B$. Indeed if we have $a\in A$ we get a function $f:B \to \{0,1\}$ defined by $b \mapsto F(a,b)$. This function is continuous and thus constant because $B$ is connected.
In the exact same way, we can show that $F$ is constant on the sets of the form $A \times \{b\}$.
We now show that this implies that $F$ is constant on $A\times B$. Indeed fix $(a,b) \in A \times B$. Now let's consider another point $(a',b')\in A \times B$. By what we have done earlier we have $F(a,b)=F(a,b')=F(a',b')$.
We are done.
A: Theorem. If $\{X_i\}_{i\in I}$ is a family of connected spaces such that $\bigcap_{i\in I} X_i\neq \emptyset$ then $\bigcup_{i\in I} X_i$ is connected.
Using this it is easy to prove what you want:
Fix $y\in Y$ and consider for every $x\in X$ the set, $$U_x=(\{x\}\times Y)\cup (X\times \{y\}).$$ Then every $U_x$ is connected for it is union of connected sets ($\{x\}\times Y\simeq Y$ and $X\times \{y\}\simeq X$) with non-empty intersection ($(\{x\}\times Y)\cap (X\times \{y\})=(x, y)$). It is easy to see $$X\times Y=\bigcup_{x\in X} U_x,$$ and since $\displaystyle \bigcap_{x\in X} U_x=X\times \{y\}\neq \phi$, $X\times Y$ is connected.
A: Suppose $U , V \subseteq A \times B$ are disjoint open sets whose union is all of $A \times B$. Fixing some $b \in B$, note that the subspace $A \times \{ b \}$ of $A \times B$ is homeomorphic to $A$, and $A \times \{ b \} \subseteq U \cup V$.  By the connectedness $A$ (and hence of $A \times \{ b \}$) we may conclude, without loss of generality, that $A \times \{ b \} \subseteq U$. 
Now given $a \in A$, knowing that $\langle a , b \rangle \in U$ go through a similar argument as above to conclude that $\{ a \} \times B \subseteq U$.
A: Hint: Try to write a continuous function $f:A\times B\to \{0,1\}$ and recall that a topological space is connected if and only if every continuous function from it to $\{0,1\}$ is constant.
A: Suppose $A,B$ are connected. To show that $A\times B$ is also connected, it suffices to show that any continuous map $f: A\times B\to \{0,1\}$ is constant. Let's do this right away.
Consider some continuous map $f: A\times B\to \{0,1\}$, and fix $(a_1,b_1) \in A\times B$. Now define $g: A\to\{0,1\}$ so that $g:a\mapsto f(a,b_1)$, and $h:B\to\{0,1\}$ so that $h: b\mapsto f(a_1,b)$. Since $A,B$ are connected, $g$ and $h$ must be necessarily constant. So, $f(a_1,b) = f(a,b_1) = f(a_1,b_1)$ for all $(a,b)\in A\times B$.
Now, fix some other $(a_2,b_2)\in A\times B$, i.e. $(a_1,b_1) \ne (a_2,b_2)$. Repeat the same process as above to obtain $f(a_2,b) = f(a,b_2) = f(a_2,b_2)$ for all $(a,b)\in A\times B$.
In $f(a_1,b) = f(a_1,b_1)$, put $b = b_2$ to get $f(a_1,b_1) = f(a_1,b_2)$. In $f(a,b_2) = f(a_2,b_2)$ put $a = a_1$ to get $f(a_2,b_2) = f(a_1,b_2)$.
This gives $$f(a_1,b_1) = f(a_2,b_2)$$
The choice of $(a_1,b_1)$ and $(a_2,b_2)$ was arbitrary, so indeed $f(a_1,b_1) = f(a_2,b_2)$ for all $(a_1,b_1), (a_2,b_2) \in A\times B$. $f$ is a constant function. Done! $\ \ \blacksquare$
This can be generalized! If $A_1,A_2,\ldots,A_n$ are connected, then so is $A_1\times A_2\times \ldots\times A_n$. You can see this in at least two ways. One, try using induction on $n$, and the result we proved above for $n=2$. $n=2$ is also the base case of induction. Alternatively, you may prove this from scratch by a similar procedure, i.e. selecting $(a_1^{(k)},a_2^{(k)},\ldots,a_n^{(k)}) \in A_1\times A_2\times \ldots\times A_n$ for $k=1,2,\ldots,n$. This is more cumbersome and might be difficult to follow, but the essence remains the same!

Food for Thought:
At this point, I wonder if this can be extended to arbitrary products - i.e. if $\{A_i\}_{i\in I}$ constitute a collection of connected sets, is $\prod_{i\in I} A_i$ also connected? (The answer is Yes).
