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In the basic analysis, we proved the following with the $\epsilon - \delta$ method.


For metric spaces $X$ and $\mathbb{R}$, $fg:X\rightarrow \mathbb{R}$ is continuous, if $f,g:X\rightarrow \mathbb{R}$ are continuous.


In my text book (Munkres, Topology), I found that the author uses the above property when $X$ is a (normal) space. (For the proof of the Tieteze extension into $\mathbb{R}$; $\forall$ A clsoed subset $A$ of a normal space $X$, a continuous function $f:A\rightarrow \mathbb{R}$ can be continuously extended to $\bar f:X\rightarrow \mathbb{R}$ s.t. $\bar f(x)=f(x)$ $ \forall x \in A$)

Is one able to use the above property for continuous functions from an arbitrary topological space into the real field?

Any help will be appreciated. Thank you.

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    $\begingroup$ Yes, it holds for arbitrary topological spaces as the domain. The same proof works if you replace "for all $y$ such that $d(x,y) < \delta$" with "for all $y$ in the neighbourhood $U$ of $x$". Or you can just prove that $\mu\colon \mathbb{R}^2\to\mathbb{R},\; \mu(x,y) = x\cdot y$ is continuous and write $fg$ as $\mu \circ (f,g)$, where $(f,g)(x) = (f(x),g(x))$. $\endgroup$ – Daniel Fischer May 18 '15 at 17:55
  • $\begingroup$ Now I see, thank you :-) $\endgroup$ – alemonk May 18 '15 at 18:01
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Yes, this is true more generally when mapping any space into any topological group $G$, since the multiplication map $ \mu : G \times G \to G$ is continuous by definition. Given maps $f, g: X \to G$ their product $f \cdot g : X \to G$ is just the composite of the pairing $(f, g) : X \to G \times G$ with the multiplication. That is, $f \cdot g = \mu \circ (f, g)$.

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