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I have the following function:

$f: \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{R}}: x\mapsto (e^{t\sin(x)})_{ t\in \mathbb{R}}$

I have to investigate if this function is continuous. I am intimidated of the space $ \mathbb{R}^{\mathbb{R}}$. Is there a general rule how to approach such problems?

I suppose I need to use topological spaces/metric spaces in order to do anything at all here. My naive attemp was at first this:

$f_1:x \mapsto \sin(x)$ is continuous. $f_2:x\mapsto e^x$ is continuous. Therefore, $f_1\circ f_2$ is also continuous, since I know this from general topology. But I have a set of Functions here and I need somehow to use the product topology...

Any help on this?

Averroes2
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1 Answers1

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For spaces $X, Y$, there are many possible topologies for $Y^X$ (as interpreted as $Y^X = \mathcal C (X,Y)$, the space of continuous maps $X → Y$), the two most natural being

  • the compact-open topology, which is the coarsest topology such that all sets of functions are open that map some fixed compact set of $X$ to some fixed open set of $Y$, and
  • the product topology, which is the finest topology making all projections – in this case interpretable as evaluations – $Y^X → Y,~h ↦ h(x)$ for $x ∈ X$ continuous.

For the compact-open topology. For topological spaces $X$, $Y$ and $Z$ with $Y$ being locally compact, there turns out to be a bijection called the exponential law, namely $$Z^{X×Y} \to (Z^Y)^X$$ This bijection is also called currying.

In this case, the uncurried version of your $f$ is the map $$ℝ × ℝ → ℝ,~(x,t) ↦ \mathrm{e}^{t \sin x}.$$ Now the exponential law implies that $f$ is continuous if and only if this map is – so is it?

For the product topology, a map $f \colon X → Z^Y$ is continuous if and only if all its components, given by $X → Z,~x ↦ f(x)(y)$ for $y ∈ Y$ are.

In your case, the components are for all $t ∈ ℝ$ given by $$ℝ → ℝ,~x ↦ \mathrm{e}^{t\sin x}.$$ Are these continuous?

Caveat. As Asaf Karagila points out, $Y^X$ is ordinarily just interpreted as the set of all maps $X → Y$, in which case the single most natural topology is again the product topology, once more defined as the finest topology making all projections $Y^X → Y,~h ↦ h(x)$ for $x ∈ X$ continuous. In this case, the discussion about the product topology still holds and you can the check the continuity of $f$ component-wise.

k.stm
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  • The components are as I said in my question continuous, being $e^x$ and $\sin(x)$. Is this what you mean? – Averroes2 Sep 24 '20 at 06:02
  • @Averroes2 No. For a map $f \colon X → \prod_{i ∈ I} Z_i$, its components are the maps $f_i \colon X → Z$ for $i ∈ I$ such that for $x ∈ X$, we have $f(x) = (f_i(x)){i ∈ I}$. Now in your case all $Z_i = ℝ$ and $I = ℝ$, making $\prod{i ∈ I} Z_i = ℝ^ℝ$, set-theoretically. More specifically, we have in your case for $X = ℝ$, $Y = ℝ$ and $Z = ℝ$ the maps $ℝ → ℝ,~x ↦ \mathrm{e}^{t \sin x}$ for $t ∈ ℝ$ as components of your map $f \colon ℝ → ℝ^ℝ$. – k.stm Sep 24 '20 at 06:06
  • Please note that I just cleared some mix-ups in the notations. I hope all is right now, but I have to leave urgently and can’t check for a few hours. – k.stm Sep 24 '20 at 06:11
  • Ok, got it. So the problem results in showing that $e^{t\sin(x)}$ is continuous. And this is continuous as a product of continuous functions. Is this right? – Averroes2 Sep 24 '20 at 06:13
  • I've never seen $X^Y$ interpreted as $\mathcal C(Y,X)$, with the exception of somewhat trivial scenarios, or when explicitly working in a category theoretic context where $X^Y$ is the set of "relevant morphisms", which in this case would indeed be the continuous functions. When someone wants to talk about $\mathcal C(Y,X)$, they just use that notation. Of course, if you happen to be the teacher of this user, and you know that $X^Y$ means just that, this is fine. But it is extremely confusing for the rest of us, since no such context was set. – Asaf Karagila Sep 24 '20 at 14:01
  • @AsafKaragila I hope this wasn’t too confusing as I explicitly explained this usage in the first sentence. I take it that the major point of your comment is that I went too far in interpreting the given question, which for one seems to invalidate my answer and for another suggests that this notational interpretation is standard when it is in fact not. Well, that’s a good point and so anyway, I have added a caveat to that effect. Hope this suits you. – k.stm Sep 24 '20 at 14:25
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    Hm, I have seen $X^Y$ used for the continuous functions (sometimes even itself given the compact-open topology) $Y \to X$ before. – Qi Zhu Sep 24 '20 at 15:52