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A finite cartesian product of connected spaces is connected.

This proposition is given as True in the topology textbook.

I'd like to prove it via contradiction.

First, let's suppose those space(finite cartesian product of connected spaces) is disconnected, then there exists a separation which halves the space into two disjoint sets.

Then for this disjointed sets, in which way should I use the connectedness to derive a contradiction?

  • Additionally, is there any infinite cartesian product of connected spaces that is disconnected?
snapper
  • 589
  • Look here https://math.stackexchange.com/questions/338027/product-of-connected-spaces and here https://math.stackexchange.com/questions/1322057/an-arbitrary-product-of-connected-spaces-is-connected?rq=1 – studiosus Apr 06 '18 at 07:37
  • For a family ${S_i}{i\in I}$ of spaces, he weakest topology on $P=\prod{i\in I}S_i$ such that each projection $p_i:P\to S_i $ is continuous, is called the Tychonoff product topology. Because it has been so influential it is sometimes called "the" product topology. The box product topology is stronger ( It equals the Tychonoff when $I$ is finite) and for some classes of spaces (e.g. metric spaces) there are other product topologies (e.g. the uniform topology). – DanielWainfleet Apr 06 '18 at 08:34

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