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The sum and product of two continuous functions is continuous

I can prove this easily when the space is metrizable, but I don't get it when the space is non-metrizable. Is there a counterexample of this? or it is true for all topological spaces those have a binary operation?

edit: It it true even if the space has a non-contionuous binary operation?

Jay C. Lee
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    The proof is basically the same : it relies on the fact that $+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ and $\times :\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ are continuous. – Captain Lama Apr 25 '16 at 10:47
  • I don't get it why the summation and product are continuous even though the space is arbitrary. – Jay C. Lee Apr 25 '16 at 11:01
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    The space is not arbitrary at all, it's $\mathbb{R}$ (of course you can replace it with a topological ring). – Captain Lama Apr 25 '16 at 11:02
  • my question is it is true for all topological spaces those have a binary operation. – Jay C. Lee Apr 25 '16 at 11:05
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    Almost. It's true for all topological spaces that have a continuous binary operation. Then use the same proof as with $\mathbb{R}$. – Captain Lama Apr 25 '16 at 11:06
  • Though you've got a good answer, you might consider making your question precise for the benefit of future readers. The comments indicate you're implicitly asking about $X$-valued maps on $X$, while the wording strongly suggests you mean real-valued functions. – Andrew D. Hwang Apr 25 '16 at 12:15

1 Answers1

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If $X$ is any space and $f,g: X \to \mathbb R$ are any two continuous functions, then $(f,g): X \to \mathbb R \times \mathbb R$ is continuous by definition (or construction) of the product, and $+: \mathbb R^2 \to \mathbb R$ is a continuous function. Then $f+g = {+} \circ (f,g)$, and a composition of continuous functions is continuous.

Similarly for $\times$, of course.

Edit: or replace $\mathbb R$ with $\mathbb C,\mathbb Q$, or any topological ring -- that is, a ring so that $+,-,\times$ are continuous.

Mees de Vries
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  • how about over an arbitrary semigroup with topology? – Jay C. Lee Apr 25 '16 at 10:55
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    @Jay: As Captain Lama already said above, the binary operation under consideration has to be continuous. In fact, if you have any topological space $X$ with a noncontinuous binary operation $: X \times X \to X$ then you can find two continuous functions into $X$ whose $$ is not continuous; namely, just take the two projections $X \times X \to X$. – Mees de Vries Apr 25 '16 at 11:18
  • Thank you. That's the answer that I wanted. – Jay C. Lee Apr 25 '16 at 11:20