Tychonoff's theorem for $[0,1]^\mathbb{R}$

Question

According to Tychonoff's theorem any uncountable product of compact spaces is compact with respect to product topology.

Then $[0,1]^\mathbb{R}$, the space of all functions defined on $\mathbb{R}$ taking values in $[0,1]$ is compact w.r.t. the product topology.

Consider the function $\delta_0(x)=\max(0,\min(x,1))$ on real numbers and $\delta_t(x)=\delta_0(x-t)$. For $t\to\infty$, it seems that there is no convergent sub-sequence and again it seems that $[0,1]^\mathbb{R}$ is then not compact or sequentially compact?

Could someone point out the problem I have? Thanks.

score 4 · Accepted Answer · answered Mar 25 '15 at 22:08

4

The product topology is the topology of pointwise convergence; your functions converge pointwise to the zero function.

You are correct that $[0, 1]^{\mathbb{R}}$ is not sequentially compact; I believe $f_n(x) = |\sin nx|$ is an explicit counterexample, but I haven't checked it carefully. But neither sequential compactness nor compactness imply the other in general (and in particular $[0, 1]^{\mathbb{R}}$ is not metrizable).

answered Mar 25 '15 at 22:08

Qiaochu Yuan

419,620

thx for the answer. yes your example is quite okay. – Seyhmus Güngören Mar 25 '15 at 22:21

Moya · Answer 2 · 2015-03-25T22:21:18.530

1

See Munkres Topology section 21 example 2 (page 133). This proves that the set $\mathbb{R}^{\mathbb{R}}$ is not metrizable in the product topology. In particular, $[0,1]^{\mathbb{R}}$ is not metrizable, although Tychonoff tells us that it is compact. Compactness and sequential compactness are equivalent for metric spaces. Qiaochu Yuan provides a proper counterexample.

Edit: Section 28 of the same reference proves this point.

edited Mar 25 '15 at 22:21

answered Mar 25 '15 at 22:10

Moya

5,248

interesting. thx. – Seyhmus Güngören Mar 25 '15 at 22:20

Tychonoff's theorem for $[0,1]^\mathbb{R}$

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