I have been playing around with the idea of thinking of the set of real valued functions $\mathbb{R}^{\mathbb{R}}$ as the set of uncountably long tuples in $\mathbb{R}$-dimensional space, and I have a question which I have had trouble answering for myself which I would like some help with.

I was wondering how one could construct an explicit bijection from $\mathbb{R}^{\mathbb{R}} \to \mathbb{R}^{(a,b)}$ (for $a,b\in \mathbb{R}$) since $|(a,b)|=|\mathbb{R}|$. My intuition tells me that the bijection should be defined in terms of the range of each function (since that is what distinguishes them), but I have had trouble determining an explicit formula. Also, are there any other interesting bijections that we can obtain, or homeomorphisms if we equip $\mathbb{R}^{\mathbb{R}}$ with the metric $d(f,g)=sup_{x\in \mathbb{R}} \frac{d(fx, gx)}{1+d(fx, gx)}$?

I think it would be interesting to see what we could continuously deform the space of real valued functions (or a subspace of it) into, but I'm having trouble constructing a bijection (let alone a bicontinuous one) to get me started.

Thanks in advance


1 Answer 1


Let $f: (a,b) \rightarrow \mathbb{R}$ be a bijection. Then,

$$\mathbb{R}^{\mathbb{R}} \rightarrow \mathbb{R}^{(a,b)}$$ $$g \mapsto g \circ f$$

is a bijection, with inverse being

$$\mathbb{R}^{(a,b)} \rightarrow \mathbb{R}^{\mathbb{R}}$$ $$g \mapsto g \circ f^{-1}.$$

As for a homeomorphism, note that this is an isometry.

Why it's an isometry:

Take two functions $\xi, \eta : \mathbb{R} \rightarrow \mathbb{R}$. Denote by $d_1$ the distance in $\mathbb{R}^{\mathbb{R}}$ and $d_2$ the distance in $\mathbb{R}^{(a,b)}$. Then, we have that $$d_1(\xi,\eta)=\sup_{x \in \mathbb{R}} \frac{d(\xi(x),\eta(x))}{1+d(\xi(x),\eta(x))}=\sup_{x \in (a,b)} \frac{d(\xi(f(x)),\eta(f(x)))}{1+d(\xi(f(x)),\eta(f(x)))}=d_2(\xi \circ f, \eta \circ f),$$ where the middle equality holds due to the fact that $f$ is a bijection.

  • $\begingroup$ Would you mind editing the post to elaborate on the isometry aspect of it? (please and thank you). $\endgroup$ Dec 30, 2015 at 7:14
  • $\begingroup$ @SirJective I edited. Hope it clarifies : ) $\endgroup$
    – Aloizio Macedo
    Dec 30, 2015 at 19:04
  • $\begingroup$ Yes it does thank you very much. Do you know of any results regard equivalences between spaces like these? Is there something familiar that we prefer to think of $\mathbb{R}^{\mathbb{R}}$ as? $\endgroup$ Dec 30, 2015 at 20:11

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