Tietze extension theorem for $X=Y=\mathbb{R}^\omega$

descriptive-set-theoryfunctional-analysisgeneral-topology

The Tietze extension theorem says that if $X$ is a Polish space (even a normal space) and $Y=\mathbb{R}^n$, then a continuous function $f:C \rightarrow Y$ on a closed set $C \subseteq X$ can be extended to a continuous function $g:X \rightarrow Y$.

My questions are:

  1. If we consider $X=Y=\mathbb{R}^\omega$, i.e. the countable product of $\mathbb{R}$ which is universal for Polish spaces, does the theorem still hold?
  2. If not, do we have a counter-example?

Thanks!

Best Answer

Let $X$ be any normal space. You know that the Tietze extension theorem is true for $Y = \mathbb R$. Now let $f : C \to \mathbb{R}^\omega$ be map defined on a closed $C \subset X$. Letting denote $p_n : \mathbb{R}^\omega \to \mathbb R$ the projection onto the factor with index $n$, all $p_nf : C \to \mathbb R$ are continuous and have a continuous extension $g_n : X \to \mathbb R$. The map $g : X \to \mathbb{R}^\omega, g(x) = (g_n(x))$, is continuous by the universal property of the product. It is an extension of $f$.

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