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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?

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  • $\begingroup$ You need to use the product topology on $ℝ^ℝ$? $\endgroup$
    – k.stm
    Sep 24, 2020 at 5:37
  • $\begingroup$ I don't need to, but I think this has something to do with projections... $\endgroup$
    – Averroes2
    Sep 24, 2020 at 5:38
  • $\begingroup$ With which topology is $ℝ^ℝ$ equipped? $\endgroup$
    – k.stm
    Sep 24, 2020 at 5:39
  • $\begingroup$ The problem says nothing about this... $\endgroup$
    – Averroes2
    Sep 24, 2020 at 5:39

1 Answer 1

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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.

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  • $\begingroup$ The components are as I said in my question continuous, being $e^x$ and $\sin(x)$. Is this what you mean? $\endgroup$
    – Averroes2
    Sep 24, 2020 at 6:02
  • $\begingroup$ @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 $f \colon ℝ → ℝ^ℝ$. $\endgroup$
    – k.stm
    Sep 24, 2020 at 6:06
  • $\begingroup$ 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. $\endgroup$
    – k.stm
    Sep 24, 2020 at 6:11
  • $\begingroup$ 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? $\endgroup$
    – Averroes2
    Sep 24, 2020 at 6:13
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    $\begingroup$ Hm, I have seen $X^Y$ used for the continuous functions (sometimes even itself given the compact-open topology) $Y \to X$ before. $\endgroup$
    – Qi Zhu
    Sep 24, 2020 at 15:52

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