I will show that for all $ i\in I$ (finite), $X_{i}$ is connected implies that $\prod_{i\in I} X_{i}$ is connected.

I have already shown that $X_{1}$ and $X_{2}$ are connected.

Now I am supposed to use the mathematical induction, but I am not sure how to do.

How can we show the following:

Suppose that $\prod_{i=1}^{n} X_{i}$ is connected. Is $\prod_{i=1}^{n+1}X_{i}$ also connected?

  • $\begingroup$ Your phrasing is somewhat confusing. Are you assuming each $X_i$ is connected? $\endgroup$ – Aweygan Apr 16 '18 at 4:19
  • $\begingroup$ Yes, I am editting it right now $\endgroup$ – Salih Akın Apr 16 '18 at 4:20

If you can show that the product of two connected spaces is connected, then the result follows. This is because, in the inductive step, $\prod_{i=1}^{n+1}X_{i}$ is homeomorphic to $\left(\prod_{i=1}^{n}X_{i}\right)\times X_{n+1}$.

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  • $\begingroup$ By the homeomorphism, if $(\prod_{i=1}^{n}X_{i})\times X_{n+1}$ is connected, then $\prod_{i=1}^{n+1}X_{i}$ will be connected. But to say this, we first have to say $X_{n+1}$ is connected? $\endgroup$ – Salih Akın Apr 16 '18 at 4:39
  • $\begingroup$ Yes, we need $X_{n+1}$ is connected. $\endgroup$ – Aweygan Apr 16 '18 at 4:40
  • $\begingroup$ But we don't know that $\endgroup$ – Salih Akın Apr 16 '18 at 4:41
  • $\begingroup$ Yes we do. We are assuming that $X_i$ is connected for all $i$. Then in the inductive step, we get the additional hypothesis that $\prod_{i=1}^{n}X_{i}$ is connected. $\endgroup$ – Aweygan Apr 16 '18 at 4:43
  • $\begingroup$ I understand now, thank you! $\endgroup$ – Salih Akın Apr 16 '18 at 4:45

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