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I would like to show that a countable product of sequentially compact spaces is a sequentially compact space.

To show that a product of N sequentially compact spaces is sequentially compact, I think that we can take a sequence $(x_1, x_2, \cdots, x_N)$ in the product of sequentially compact spaces, and say that each sequence $x_1, x_2, \cdots, x_N$ is in the sequentially compact space $X_i$, so has a convergent subsequence. Then the global sequence converges to the sequence of limits of the sequences $x_1, x_2, \cdots, x_N$.

I don’t understand why this reasoning would not work in the case of an infinite product space and how to prove, then, the result.


marked as duplicate by Henno Brandsma general-topology Feb 28 at 22:30

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  • 3
    $\begingroup$ Diagonalize. See this. $\endgroup$ – David Mitra Feb 28 at 8:23
  • $\begingroup$ Thank you but I don’t undestand the answer given 😢 it seems to be more an explanation of the principle of diagonalizing than a proof $\endgroup$ – user637846 Feb 28 at 8:30