Prove that the metric spaces $(X,d)$ and $(X,d(x,y)^{1/3})$ are homemorphic. Let $(X,d)$ be a metric space, where $d$ is the standard metric. Define a new metric $d': \mathbb{R} 
\times \mathbb{R} \rightarrow$ by
$$ d'(x,y) = d(x,y)^{1/3}. $$
Prove that $(X,d)$ and $(X,d')$ are homemorphic.
I'm very unsure on how to prove this, but I am assuming that the idea is to show that one can construct a function $f: (X,d) \rightarrow (X,d')$ that is continuous, bijective and has a continuous inverse. But without a specific choice for $f$, I do not know how to prove this. If I am on the right track, how does one pick an appropriate $f$? (This is not homework)
 A: Here's another simple way to view it: actually, the topologies $\mathcal{T}$ and $\mathcal{T}'$ induced in $X$ by $d$ and $d'$, respectively, are actually the same topology.
Indeed, $U$ is open in $\mathcal{T}$ $\iff$ for all $x\in U$ there exists $\varepsilon>0$ such that $\mathrm{B}_d(x,\varepsilon)=\mathrm{B}_{d'}(x,\varepsilon^{1/3})\subset U$ (where $\mathrm{B}_d$ and $\mathrm{B}_{d'}$ denote open balls for the metrics $d$ and $d'$, respectively) $\iff$ $U$ is open in $\mathcal{T}'$.
A: Just take $f(x)=x$ (that is, $f$ is the identity function).
If $x_0\in X$, and $\varepsilon>0$, if you take $\delta=\sqrt[3]\varepsilon$, then\begin{align}d'(x,x_0)<\delta&\iff\sqrt[3]{d(x,y)}<\sqrt[3]\varepsilon\\&\iff d(x,y)<\varepsilon.\end{align}This proves that $f$ is continuous as a map from $(X,d')$ onto $(X,d)$. Can you do it in the other direction?
A: There's actually a really nice way to solve the problem by appealing to the sequential definition of continuity. Let $f: (X,d) \to (X,d')$ be the identity mapping. Let $x_0 \in X$ be a fixed point and let $(a_n)_{n \in \mathbb{N}}$ be a sequence that converges to $x_0$ in the metric $d$. Then:
$$\lim_{n \to \infty} d(a_n,x_0) = 0$$
But using the continuity of the cube root function, we have that:
$$\lim_{n \to \infty} (d(f(a_n),f(x_0)))^{\frac{1}{3}} = (\lim_{n \to \infty} d(a_n,x_0))^{\frac{1}{3}} = 0$$
Now, we want to prove continuity of the inverse of $f$. Once again, let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $X$ that converges to $x_0$ in the metric $d'$. So:
$$\lim_{n \to \infty} d'(a_n,x_0) = 0$$
But now, we use the continuity of the cubic $x \mapsto x^3$ to conclude that:
$$\lim_{n \to \infty} d(f(a_n),f(x_0)) = \lim_{n \to \infty} (d'(a_n,x_0))^3 =(\lim_{n \to \infty} d'(a_n,x_0))^3 = 0$$
Hence, $f$ is a homeomorphism and you're done.
