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My professor taught us on continuity of functions in metric spaces and Homeomorphism. He defined the homeomophism as the following:

Let ($X, d$) and ($X', d'$) be metric spaces. A function $f : X \to X'$ is homeomorphism if

i. $f$ is bijection

ii. $f$ is continuous

iii. $f^-$$^1$ is also continuous

$\\$

A homeomorphism $f : X \to X'$ is an isometry if:

$d(x_1, x_2)$ = $d(f(x_1), f(x_2))$, $\forall x_1, x_2 \in X$

He then asked us to solve the following questions:

We consider isometries from $\mathbb R$ to itself with the usual metric.

Q1. Is $f(x) = x^3$ a homeomorphism? an isometry?

Q2. Is $f(x) = x + sin (x)$ a homeomorphism? an isometry?

I am confused on the last part iii. When it says $f^-$$^1$ is continuous, does it mean $f^-$$^1$ continuous on $X'$?

For Q1, I had no problem in proving bijection and continuity of $f(x)$ but kind of got stuck at $f^-$$^1$.

For Q2, Again, the bijection part is easy to show and I can see that this is a strictly increasing function but how do I show that this function is continuous?. I don't think there is simple way to express the inverse function of $x + sin(x)$. I find it hard to visualize it in my head. I don't know how to approach this on the inverse function part.

Any hint/suggestion will be appreciated.

(I'd like to solve these problems in terms of $\epsilon$ - $\delta$, i.e. $f$ is continuous at $x_o$ $\forall \epsilon > 0, \exists\delta > 0$ such that if |$x - x_o$| < $\delta$, then |$f(x) - f(x_o)$| $< \epsilon$ )

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    Answers: inverse of $x^{3}$ is $x^{1/3}$ which is continuous. The function in Q2 is a homeomorphism because it is a strictly increasing continuous bijection of $\mathbb R$. It is not an isometry: consider the points $0$ and $\pi /2$. – Kavi Rama Murthy Nov 25 '18 at 12:05

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Consider$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&x^3,\end{array}$$which is continuous. Then $f^{-1}$ is the map$$\begin{array}{rccc}f^{-1}\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\sqrt[3]x,\end{array}$$which is continuous too. Therefore, $f$ is a homeomorphism.

Now, consider$$\begin{array}{rccc}g\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&x+\sin(x),\end{array}$$which is continuous too. Since $\lim_{x\to\pm\infty}g(x)=\pm\infty$, $g$ is surjective. And, since $g'(x)>0$ except at the points of $\pi\mathbb Z$, $g$ is strictly increasing, and therefore injective. So, $g$ is a bijection. It turns out that the inverse of every continuous bijection from an interval of $\mathbb R$ onto another interval of $\mathbb R$ is continuous too. Therefore, $g$ is also a homeomorphism.

But neither $f$ nor $g$ are isometries. It's easy to find examples of points $x,y\in\mathbb R$ such that$$\bigl\rvert f(x)-f(y)\bigr\rvert,\bigl\rvert g(x)-g(y)\bigr\rvert\neq\lvert x-y\rvert.$$

José Carlos Santos
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