Solution
Let A and B be connected spaces such that a map
F:A×B"\\to" {0,1}
To prove that A×B is connected we have to show that F is a constant.
Let
"(a_1,b_1)\\in A\u00d7B"
Now ,we define
g:A"\\to" {0,1}
g:a"\\to" f(a,b1)
and
h:B"\\to" {0,1}
h:b"\\to" f(a1,b)
Since A and B are connected, h and g must be constant, hence
f(a,b1)=f(a1,b)=f(a1,b1) for all (a,b)"\\in" A×B
If we fix some other;
"(a_2,b_2)\\in A\u00d7B"
But "(a_1,b_1)\\ne (a_2,b_2)"
We obtain
f(a,b2)=f(a2,b)=f(a2,b2) for all (a,b)"\\in" A×B
In
f(a1,b)=f(a1,b1), take b=b2 ,we have
f(a1,b1)=f(a1,b2)
In
f(a,b2)=f(a2,b2), take a=a1, we have
f(a1,b2)=f(a2,b2)
Hence
f(a1,b1)=f(a2,b2)
Since
(a1,b1) and (a2,b2) are arbitrary;
f(a1,b1)=f(a2,b2) for all (a1,b1),(a2,b2) "\\in" A×B
Therefore, F is a constant function.
Hence A×B is connected.
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