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

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?

• You need to use the product topology on $ℝ^ℝ$? Sep 24, 2020 at 5:37
• I don't need to, but I think this has something to do with projections... Sep 24, 2020 at 5:38
• With which topology is $ℝ^ℝ$ equipped? Sep 24, 2020 at 5:39

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.

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