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Let $S^n$ be the n-sphere with respect to $d_2$ metric (standart metric). Let $C^n$ be the n-sphere with respect to $d_1$ metric. Clearly we have,

$S^n$ $\cong$ $C^n$ (homeomorphism).

Yet, can one argue this is true because $d_1$ and $d_2$ are equivalent metrics? Would this idea generalize?

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  • $\begingroup$ What definition of equivalence of metrics are you using? $\endgroup$ – YuiTo Cheng May 6 at 13:30
  • $\begingroup$ I'm confused what the question is. You want to show that $S^n$ is homeomorphic to itself with various metrics? $\endgroup$ – Elliot G May 6 at 13:44
  • $\begingroup$ @YuiToCheng AFAIK all common equivalence definitions imply topological equivalence? $\endgroup$ – freakish May 6 at 14:08
  • $\begingroup$ @freakish Of course. I'm just confirming whether this question is a duplicate of Equivalent metrics determine the same topology (there are also strong equivalences of metrics, you know...) $\endgroup$ – YuiTo Cheng May 6 at 14:10
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    $\begingroup$ @ErotemeObelus You define "strongly equivalent". Ordinary equivalence is much weaker. $\endgroup$ – Paul Frost May 6 at 15:36
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It is not a consequence of the equivalence of metrics. The reason it that the metrics $d_i$ on $\mathbb R^{n+1}$ are induced by norms $\lVert - \rVert_i$ on $\mathbb R^{n+1}$. It is well-known that all norms on $\mathbb R^{n+1}$ are equivalent, i.e. generate the same topology (the Euclidean topology) on $\mathbb R^{n+1}$. Hence each norm $\lVert - \rVert : \mathbb R^{n+1} \to \mathbb R$ is a continuous function with respect to this topology, and that is all we need to know.

So let us consider arbitrary norms $\lVert - \rVert_i$, i.e. $\lVert - \rVert_2$ is not necessarily the Euclidean norm $\sqrt{\sum_{i=1}^{n+1} x_i^2}$ and $\lVert - \rVert_1$ is not necessarily the norm $\sum_{i=1}^{n+1} \lvert x_i \rvert$. Let $C_i = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_i = 1 \}$ be the unit sphere with respect to $\lVert - \rVert_i$.

Define $h_1 : C_1 \to C_2, h_1(x) = x/\lVert x \rVert_2$ and $h_2 : C_2 \to C_1, h_2(x) = x/\lVert x \rVert_1$. Then $$h_2(h_1(x)) = h_2(x/\lVert x \rVert_2) = \dfrac{x/\lVert x \rVert_2}{\lVert x/\lVert x \rVert_2 \rVert_1} = \dfrac{x/\lVert x \rVert_2}{(1/\lVert x \rVert_2) \lVert x \rVert_1} = x / \lVert x \rVert_1 = x$$ since $\lVert x \rVert_1 = 1$ for $x \in C_1$. Similarly $h_1 \circ h_2 = id$.

To see that the equivalence of metrics is not enough, consider the metric $d_2$ which gives you $S^n$. Define $d'_2(x,y) = \min(d_2(x,y), 1)$. This is an equivalent metric, but $\{ x \in \mathbb R^{n+1} \mid d'_2(0,x) = 1 \} = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_2 \ge 1 \}$ which is not homeomorphic to $S^n$.

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  • $\begingroup$ would this work for any 2 equivalent norms on an arbitrary space? How did we use the equivalance of the norms? It also seems like we can generalize this to the sets containing points of some other fixed norm. $\endgroup$ – Baran Zadeoglu May 6 at 15:07
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    $\begingroup$ Yes. Equivalent norms $\lVert - \rVert_i$ on $X$ induce the same topology and the functions $\lVert - \rVert_i : X \to \mathbb R$ are continuous with respect to this topology. This is needed to see that the above functions $h_i$ are continuous. $\endgroup$ – Paul Frost May 6 at 15:31

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