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I was given this proof but I don't clearly understand it. Would someone be able to dumb it down for me so I can maybe process it better?

Since a topological space is connected if and only if every function from it to $\lbrace 0,1\rbrace$ is constant. Let $F:X\times Y\rightarrow\lbrace 0,1\rbrace$ be a continuous function. Let $x \in X$, we get a function $f:Y\rightarrow\lbrace 0,1\rbrace$ defined by $y\mapsto F(x,y)$. F is constant on every set $\lbrace x\rbrace\times Y$. So the function is continuous and constant because $Y$ is connected. In the same way $F$ is constant on the set of the form $X\times \lbrace y\rbrace$. This implies constant on $X\times Y$. $(x,y)$ exist on $X\times Y$. Say there exists $(a,b)$ on $X\times Y{}{}$. $F(x,y)=F(a,b)$ proving $A\times B$ is conected.

The main things I don't understand is why is F constant on every set $\{x\}\times Y$ and how is Y connected to make the function continiuous and constant

Martin Sleziak
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bob
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    Which part of the proof you don't understand? – pritam May 01 '13 at 09:33
  • Things I don't understand: Why are the functions $f\colon Y\to{0,1}$ constant if we are not given (are we?) that $Y$ is connected? And where do $A,B$ in the last senetcne come from? - Otherwise, the one step missing in the proof seems to be $F(a,b)=F(x,b)$ because $X\to{0,1}, t\mapsto F(t,b)$ is constant, $F(x,b)=F(x,y)$ because $Y\to{0,1}, t\mapsto F(x,t)$ is constant, hence $F(a,b)=F(x,b)=F(x,y)$; as $(x,y)\in X\times Y$ was arbitrary, $F$ is constant. – Hagen von Eitzen May 01 '13 at 09:45
  • I don't understand the whole thing. I asked my professor and I din't even understand how she explained it – bob May 01 '13 at 09:53
  • The main things I don't understand is why is F constant on every set {x}xY and how is Y connected to make the function continiuous and constant – bob May 01 '13 at 10:13
  • If a topological space $Y$ is connected, then every continuous function $f:Y\rightarrow\lbrace a,b\rbrace$ is constant, where $\lbrace a,b\rbrace$ is any discrete $2$-point space. By fixing a point $x\in X$, the function $F$ can be treated as a function from $Y$ to $\lbrace 0,1\rbrace$. Since this function is continuous on $X\times Y$, it's still continuous when we fix $x\in X$. And because $Y$ is connected, any continuous function from $Y$ to a $2$-point space must be constant. – Warren Moore May 01 '13 at 10:35
  • possible duplicate of Product of connected spaces – Najib Idrissi Apr 28 '15 at 12:01

1 Answers1

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The basic fact we use is the one you start out with (which I won't prove, as I assume it's already known; it's not hard anyway):

(1) $X$ is connected iff every continuous function $f: X \rightarrow \{0,1\}$ (the latter space in the discrete topology) is constant.

So, given connected spaces $X$ and $Y$, we start with an arbitrary continuous function $F: X \times Y \rightarrow \{0,1\}$ and we want to show it is constant. This will show that $X \times Y$ is connected by (1).

First, for a fixed $x \in X$, we can define $F_x: Y \rightarrow \{0,1\}$ by $F_x(y) = F(x,y)$.
This is the composition of the maps that sends $y$ to $(x,y)$ (for this fixed $x$) and $F$, and as both these maps are continuous, so is $F_x$, for every $x$. So, as this is a continuous function from $Y$ to $\{0,1\}$ and $Y$ is connected, every $F_x$ is constant, say all its values are $c_x \in \{0,1\}$.

Of course we haven't used the connectedness of $X$ yet, so we do the exact same thing fixing $y$: for every $y \in Y$, we define $F^y: X \rightarrow \{0,1\}$ by $F^y(x) = F(x,y)$. Again, this is a composition of the map sending $x$ to $(x,y)$ (for this fixed $y$) and $F$, so every $F^y$ is continuous from the connected $X$ to $\{0,1\}$ and so $F^y$ is constant with value $c'_y \in \{0,1\}$, say.

The claim now is that $F$ is constant: let $(a,b)$ and ($c,d)$ be any two points in $X \times Y$. Then we also consider the point $(a,d)$ and note that: $$F(a,b) = F_a(b) = c_a = F_a(d) = F(a,d) = F^d(a) = c'_d = F^d(c) = F(c,d)\mbox{,}$$

first using that $F_a$ is constant and then that $F^d$ is constant. We basically connect any 2 points via a third using a horizontal and a vertical "line" (here via $(a,d)$) and on every "line" $F$ remains constant, so $F$ is a constant function. This concludes the proof.

As a last note: the fact that a function $x \rightarrow (x,y)$, for fixed $y$, is continuous, is easy from the definitions: a basic neighbourhood of $(x,y)$ is of the form $U \times V$, $U$ open in $X$ containing $x$, $V$ open in $Y$ containing $y$, and its inverse image under this function is just $U$.

Henno Brandsma
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