Question #217033
Prove that a countable product of connected space is connected
1
Expert's answer
2021-07-19T17:09:20-0400

Let ϕ:A×B{0,1}\phi:A×B\rightarrow \{0,1\} be a continuous function. It suffices to show that ϕ\phi is constant.


Now if we fix aAa\in A then we get a function f:B{0,1}f:B \rightarrow \{0,1\} defined by bϕ(a,b)b \mapsto \phi(a,b). Since BB is connected, it follows that ff is continuous. So ff is constant, and hence ϕ\phi is continuous on all sets of the form {a}×B\{a\}×B.


Similarly we can show that ϕ\phi is constant on all sets of the form A×{b}A×\{b\}.


Next, fix (a,b)A×B(a,b) \in A × B. Also, let (a,b)A×B(a',b') \in A × B. Then from the results above, we have ϕ(a,b)=ϕ(a,b)=ϕ(a,b).\phi(a,b)=\phi(a,b')=\phi(a',b').

Hence A×BA×B is connected.



Now suppose (Ai)(A_i) is a countable collection of connected sets for iNi\in \mathbb{N}. We show that the countable Cartesian product A1×A2×...A_1×A_2×... is also connected

Clearly, A1A_1 is connected.

Suppose A1×A2×...×AkA_1×A_2×...×A_k is connected for some kNk\in \mathbb{N}.

Let L=A1×A2×...×AkL=A_1×A_2×...×A_k. Then L×Ak+1L×A_{k+1} is connected since the Cartesian product of 2 connected spaces is also connected.

So A1×A2×...×Ak×Ak+1A_1×A_2×...×A_k×A_{k+1} is connected. Hence our assertion holds by induction.


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