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Let $X$ be a vector space of $\mathbb{R}$ and $d, d'$ two equivalent distances on $X$ derived from norms, we are asked to prove that the open balls $B_{1}^{d}(0)=\{x\in X: ||x||<1\}$ and $B_{1}^{d'}(0)=\{x\in X: ||x||'<1\}$ are homeomorphic.

In class we defined $f: B_{1}^{d}(0)\to B_{1}^{d'}(0)$ which maps $x$ to $f(x)=\frac{||x||}{||x||'}\cdot x$ when $x\neq0$ and $f(0)=0$. This is well defined because it is easy to see that if $x\in B_{1}^{d'}(0)$, then $f(x)\in B_{1}^{d'}(0)$.

The problem I have is proving that $f$ is continuous. For every $x\in B_{1}^{d'}(0)$, we have to prove that $\forall \epsilon >0, \exists \delta >0$ such that $f(B_{\delta}^{d}(x))\subseteq B_{\epsilon}^{d'}(f(x))$. In class we were told that this is true if and only if $d$ and $d'$ are equivalent, but I do not understand how this affirmation is true.

As for the rest of the exercise, since we have to prove the homeomorphism, we would have to find the inverse function and prove that it is continuous as well, but it is basically the same problem.

Edit: d and d' are equivalent if and only if there exist $\alpha, \beta \ge0$ such that for every $x,y\in X$, $\alpha \cdot d(x,y)\le d'(x,y)\le \beta \cdot d(x,y)$

NFC
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  • Perhaps you could write down a definition of what it means for $d$ and $d'$ to be equivalent? – Dan Robertson Mar 29 '18 at 22:41
  • My bad, I've just added it. – NFC Mar 30 '18 at 08:51
  • And are $d,d'$ just any metric satisfying non negativity, $d(x,y)=0\Rightarrow x=y,$ and the triangle inequality, or are they induced by some norm on the vector space (which satisfies all the laws for a norm)? – Dan Robertson Mar 30 '18 at 08:53
  • $d, d'$ are both induced by norms. I should have clarified that too, sorry. – NFC Mar 30 '18 at 08:58

2 Answers2

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In any normed linear space $X$ the open unit ball $\{x:||x||<1\}$ is homeomorphic to the whole space $X$ via the map $x \to \frac x {1-||x||}$ (the inverse of this map being $x \to \frac x {1+||x||}$). Equivalent norms give the same topology on $X$ so open balls in any two equivalent norms are homeomorphic.

Kavi Rama Murthy
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Let $Y$ be the space $X$ with topology induced by $d$ and $Y'$ with topology induced by $d'.$

Let’s first suppose that $\|\cdot\|':Y\to \Bbb R$ is continuous (we already have $\|\cdot\|:Y\to\Bbb R$ continuous). Then so long as we stay away from 0 (ie the preimage of 0 under $\|\cdot\|'$), we have that $x\mapsto \frac{\|x\|}{\|x\|'}$ is continuous. (This is because it is the composition of our continuous norms with the continuous functions $\Delta(x) = (x,x)$ and $\div:\Bbb R \times \Bbb R^*\to\Bbb R$). The final step is that if $g:Y\to\Bbb R$ is continuous then so is $h:x\mapsto g(x)\cdot x.$ To prove that step, let $\epsilon >0$ and find $\delta_1>0$ such that $g(B_{\delta_1}(x))\subset ((1-\epsilon')g(x), (1+\epsilon')g(x)),$ where $\epsilon' =\min\left(\frac12,\frac\epsilon{|g(x)|\|x\|}\right)$ (or $\frac12$ if $g(x)\|x\|=0$). Now choose $0<\delta_2<(1+\epsilon')|g(x)|.$ Then given $d(x,y)<\delta_2,$ we have $$\begin{align*} d(h(x),h(y)) &= d(g(x)x,g(y)y) \\ &\le d(g(x)x,g(y)x) + d(g(y)x,g(y)y)\\ &= |g(y)-g(x)|\|x\| + |g(y)|\|x-y\|\\ &\le \epsilon'|g(x)|\|x\| + (1+\epsilon')|g(x)|\delta_2\\ &\le 2\epsilon.\end{align*}$$

The harder part to get your head round is that $\|\cdot\|'$, which is continuous as a function $Y'\to\Bbb R,$ is also continuous as a function $Y\to\Bbb R.$ That is what the key phrase “equivalent distances” is about at the start. To prove that you need to show that $$\forall x\in X\quad\forall\epsilon>0\quad\exists\delta>0\quad\forall y\in X\quad \|x-y\|<\delta\Rightarrow|\|x\|'-\|y\|'|<\epsilon.$$

Note that I have assumed your distance functions are derived from norms which follow all the associated laws.

Dan Robertson
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