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

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. 
