TITLE: On Tychonoff's theorem
QUESTION [1 upvotes]: Let $x_n:[0,1]\rightarrow E$ a sequence of applications, where $E$ is a normed space. Suppose that for every $t\in [0,1]$ $$\left \{ x_n(t):\;  n\in \mathbb N\right \} \text{ is weakly compact.}$$

Claim: By Tychonoff's theorem, one can get that:
$$\left \{ x_n(.):\;  n\in \mathbb N\right \}=\prod_{t\in [0,1]} \left \{ x_n(t):\;  n\in \mathbb N\right \}$$
is weakly compact in $\mathcal{A}([0,1],E)$, where $\mathcal{A}([0,1],E)$ is the space of application from $[0,1]$ to $E$.

Problem: The left side of the equality is a subset of $\mathcal{A}([0,1],E)$, while the right side is a subset of $\{x_n([0,1]): n\in \mathbb N\}$!
I know the following application is bijective $$\begin{matrix}
\Phi: \mathcal{A}([0,1],E)& \rightarrow & E^{[0,1]}\\ 
x &\mapsto &  \big(x(t)\big)_{t\in [0,1]}
\end{matrix}$$
but I don't know how to use it, if its homeomorphism my problem will be obvious.

REPLY [1 votes]: The right hand side (by the definition of an infinite product) is also in $\mathcal{A}([0,1],E)$.
Your $\Phi$ is the identity. There is nothing to prove.
IMO, it's the notation $\{x_n(\cdot): n \in \Bbb N\}$ that's strange.
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