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We got 2 metric spaces in $\mathbb{R}$, and these metrics: $d_1(x,y):= |x-y|$ and $d_2(x,y):= |x^3-y^3|$. I'm asked to prove this by proving that the identity which goes from one metric space to the other is an homeomorphism. I'm not sure what is the identity they are speaking about here.

José Osorio
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  • Two metrics... where ? The real line or what? – DonAntonio Apr 05 '16 at 04:29
  • edited, yes in $\mathbb{R}$ – José Osorio Apr 05 '16 at 04:30
  • Basically you have a space $X$ and there are two metrics $d_1,d_2$ on the space. The identity map is the map from $(X,d_1)$ to $(X,d_2)$ mapping $x\mapsto x$. – Nirakar Neo Apr 05 '16 at 04:30
  • Every open $d_1$-ball with centre $a$ contains an open $d_2$-ball with centre $a$, and vice-versa, so the topologies are the same. – André Nicolas Apr 05 '16 at 04:47
  • $x^3-y^3 = (x-y)(x^2+xy+y^2)$ hence $d_2(x,y) = d_1(x,y)|x^2+xy+y^2|$. you have to prove that if for a sequence $(y_n)$, when $n \to \infty$ : $d_1(x,y_n) \to 0$ then $d_2(x,y_n) \to 0$, and conversely if $d_2(x,y_n) \to 0$ then $d_1(x,y_n) \to 0$. this proves that the topologies induced by the metrics are the same : they define the same neighborhoods and the same converging sequences – reuns Apr 05 '16 at 04:50
  • Ok maybe the problem is i don't know how to even formulate the question, i tried with a function $f(x)=x$ and prove it is homeomorphism (because im asked to prove the identity is an homeomorphism) but what has to do $f(x)=x$ with the given metrics? i feel im missing someting :/ – José Osorio Apr 05 '16 at 04:52

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The identity is the function $f:\mathbb R\to\mathbb R$ such that $f(x)=x$. Here, the first $\mathbb R$ has the $d_1$ metric, and the second $\mathbb R$ has the $d_2$ metric. We need to show that $f$ is a bijection which is continuous and has a continuous inverse. It is clearly a bijection. To prove that the identity continuous, choose $x\in\mathbb R$. Let $\epsilon>0$, and let $\delta=\min\{\epsilon/(3|x|^2+3|x|+1),1\}$. Then if $d_1(x,y)<\delta$, $|y|\leq|x|+1$, and $$d_2(f(x),f(y))=d_2(x,y)=|x^3-y^3|=|x-y||x^2+xy+y^2|$$ $$\leq|x-y|(|x|^2+|x|^2+|x|+|x|^2+2|x|+1)$$ $$<\frac{\epsilon}{3|x|^2+3|x|+1}(3|x|^2+3|x|+1)$$ $$=\epsilon.$$

Therefore the identity is continuous.

Now you need to prove that the inverse of the identity is continuous.

Plutoro
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    Looks like my definition of identity wasnt' the problem, i was misunderstanding continuity itself. Thx a lot !!!!!!!!! – José Osorio Apr 05 '16 at 05:18
  • the inverse of the function f, isn't it continuous if i say "without loss of generality"? – José Osorio Apr 05 '16 at 06:06
  • No. There are plenty of examples of functions which are continuous whose inverses are not continuous. – Plutoro Apr 05 '16 at 06:09
  • but this is an identity! – José Osorio Apr 05 '16 at 06:13
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    Sorry. When we change our concept of what distance means, even the identity may be discontinuous. An example would be $d_3 (x,y)=1$ if $x\neq y$, and $0$ if $x=y $. The identity between the regular $\mathbb R $ and $\mathbb R $ with this distance is not continuous. – Plutoro Apr 05 '16 at 06:17