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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.

Martin Sleziak
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moe
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6 Answers6

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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.

titusjgr
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jeanmfischer
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  • It's a nice proof. I'm thinking how to apply this proof to show the connectivity when it is a product topology of infinitely many connected space. Any suggestions? – Zheng Liu Sep 04 '15 at 12:13
  • Do you want a product of countable many spaces or any product? – jeanmfischer Sep 06 '15 at 10:53
  • Hi. Any product. I think this theorem holds for any product, doesn't it? – Zheng Liu Sep 07 '15 at 04:42
  • Maybe! Let $(X_i){i \in I}$ be a family of topological spaces over a set $I$, and let $i \in I$. This time the ${a}\times B$ can be a set of the form ${(x_j){j \in I -{ i} }} \times X_i$, where $(x_j)_{j \in I -{ i} }$ lies in the product where $X_i$ was removed. – jeanmfischer Sep 07 '15 at 10:17
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    http://math.stackexchange.com/questions/1322057/an-arbitrary-product-of-connected-spaces-is-connected?rq=1 – jeanmfischer Sep 07 '15 at 10:26
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    Is it obvious that $f$ is continuous? – B. Mehta Jun 02 '17 at 14:59
  • If you can show that the map $b\mapsto (a,b)$ is continuous it is ! – jeanmfischer Jun 02 '17 at 15:42
  • How do we know that for $a, a' \in A$, that we have $f(b) = F(a,b) = F(a',b)$? –  Aug 02 '19 at 18:04
  • To see that $f$ is continuous, I think it's useful to observe that $f = F|_{{a} × B}$. Perhaps the restriction of a continuous map to a connected subset is continuous? Ah we don't even need the subset to be connected of course. – Jos van Nieuwman Jan 02 '21 at 16:04
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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.

Babelfish
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PtF
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  • I've proved that $f:X\times${$y$}$\rightarrow X$ is a bijective map.How to show that $f$ is continuous?I tried it by considering any open set $O$ in $X$,but missing the idea of showing $f^{-1}(O)$ is open in $X\times ${$y$}. – Styles Nov 11 '18 at 06:27
  • Prove connectedness is a topological invariant: That is, if $X$ is connected and $Y$ is homeomorphic to $X$ then $Y$ is also connected. – PtF May 17 '19 at 19:59
  • @P.Styles As far as I have understood, if you consider discrete topologies $\mathscr {D}_1$ and $\mathscr D_2$ on $X$ and $X\times{y},$ respectively, then you can readily prove that both $f$ and $f^{-1}$ are continuous. – Usual_Learner Dec 05 '21 at 20:05
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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 $\{ a \} \times B \subseteq U$.

user642796
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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.

Arrow
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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).

stoic-santiago
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  • Fik $k\in I$, call $J = I -{i}$, fix $(a_j) \in \prod_{j\in J}A_{j}$, by the universal property of the product, there is a continous map $\alpha : A_k \to \prod_{i\in I} A_i$, where $\pi_i \circ \alpha $ is the constant map with value $a_i$ if $i \neq k$ and $\pi_k \circ \alpha = id_{A_k}$. $f \circ \alpha$ is continous and is the map $f$ where we fixed the coordinates $i \neq k$ to be the $a_j$'s. $f\circ \alpha$ must be constant by hypothesis of $A_k$ beeing connected. This then can be used to show that $f$ is constant. – jeanmfischer Mar 06 '21 at 17:42
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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.P.
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