Let $X=[0,1]^{[0,1]}=\prod_{\alpha\in [0,1]}[0,1]$ with the product topology, show that $X$ is not sequentially compact.

Question

Let $X=[0,1]^{[0,1]}=\prod_{\alpha\in [0,1]}[0,1]$ with the product topology, show that $X$ is not sequentially compact.

I know that this space is compact by tychonoff's theorem but I do not know how to prove that it is not sequentially compact, I would say it can be done using the cantor's diagonal but I do not know how, could someone help me please?, thanks.

score 2 · Accepted Answer · answered Nov 26 '17 at 22:57

Note that the "index set" itself does not matter, just its size. All other powers of $[0,1]$ with the same size index set are homeomorphic. So we could instead consider $[0,1]^I$ where $I =\{0,1\}^\mathbb{N}$ (as $|[0,1]| = 2^{\aleph_0}$), and then, as I showed in this answer we get a nice diagonalisation argument as to why $[0,1]^I$, or $\{0,1\}^I$ is not sequentially compact. We only really use of the product topology that a sequence converges in a product iff all it projections converge in their respective factor.

Let $X=[0,1]^{[0,1]}=\prod_{\alpha\in [0,1]}[0,1]$ with the product topology, show that $X$ is not sequentially compact.

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