1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $\left(\prod_{i=1}^{n}X_{i}\right)\times X_{n+1}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.