# Tietze extension theorem for complex valued functions

Why is this theorem always only stated for real valued functions, and not for complex valued functions?

Thanks.

#### Solutions Collecting From Web of "Tietze extension theorem for complex valued functions"

Since the theorem uses only the topology of $\mathbb R$ and not in any way the multiplicative structure, $\mathbb C$ is basically the same as $\mathbb R^2$. But the theorem easily generalises to $\mathbb R^n$: just extend each component separately.

The wikipedia article on Tietze’s Extension Theorem mentions that one can replace $\mathbb{R}$ with $\mathbb{R}^I$ for any index set $I$. Taking $\# I = 2$ — and, of course, using that $\mathbb{C}$ is homeomorphic to $\mathbb{R}^2$! — we get the result you are asking about. So to my mind this is a standard reference which includes the version of the theorem you are asking about.

The version for $\mathbb{C}$-valued functions follows immediately from the version for $\mathbb{R}$-valued functions, because a function $f: X \rightarrow \mathbb{C}$ is continuous iff its real and imaginary parts $\Re f, \Im f: X \rightarrow \mathbb{R}$ are continuous: this is the universal property of the direct product. So you can apply the $\mathbb{R}$-valued Tietze Extension Theorem to each component to get the $\mathbb{C}$-valued version. (This argument works verbatim with $\mathbb{R}^2$ replaced by $\mathbb{R}^I$ for any set $I$: it is again the universal property of the direct product.)