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I have a question about equivalency in metric spaces. I don't know how to Prove this. $U_1(x, r) \subseteq U_2(x, \varepsilon)$ and $U_2(x, r) \subseteq U_1(x, \varepsilon)$, where $U_j(x,r) := \{y \in X : d_j(y,x) < r\}$, $j = 1,2$. Show that $(X,d_1)$ and $(X,d_2)$ are equivalent if and only if the convergent sequences in $(X,d_1)$ and those in $(X,d_2)$ coincide. Can you help me with a hint I do Not know how to Apply convergence.

Norse
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Tobias Feynman
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  • It is not clear what the statement $U_1(x, r) \subseteq U_2(x, \varepsilon)$ and $U_2(x, r) \subseteq U_1(x, \varepsilon)$ is doing in the middle of your question. I suspect that this is meant to be your definition of equivalency. If that is the case, please make that clear in your question, and explain what exactly $r$ and $\varepsilon$ are here. – Ben Grossmann Jan 20 '20 at 19:20
  • Hey yes thats true I think r is the distance from the metric and the epsilon is the neighbourhood – Tobias Feynman Jan 20 '20 at 19:23
  • If this is a homework question, could you just copy what the question says about $r$ and $\varepsilon$ word for word? – Ben Grossmann Jan 20 '20 at 19:27
  • Twometricsd1 andd2 onasetXarecalledequivalentifforeachx∈Xandeachε>0 there exists r > 0 such that U1(x, r) ⊂ U2(x, ε) and U2(x, r) ⊂ U1(x, ε), where Uj(x,r) := {y ∈ X : dj(y,x) < r}, j = 1,2. Show that (X,d1) and (X,d2) are equivalent if and only if the convergent sequences in (X,d1) and those in (X,d2) coincide. – Tobias Feynman Jan 20 '20 at 20:57

1 Answers1

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A metric space is sequential: a set is closed iff its is sequentially closed. This already follows from first countability. So if $d_1$ and $d_2$ give the same convergent sequences, they make the same sets sequentially closed, so they have the same closed (and open) sets, and thus induce the same topology, and are equivalent.

You could also give a boring direct proof based on the given (not quite correct, I give the correct formulation here) definition plus the definition of convergence; a start:

Suppose $d_1$ and $d_2$ are equivalent, and $x_n \to x$ under $d_1$. To see that $x_n \to x$ under $d_2$, let $\varepsilon>0$. By equivalence we have for $x$ and $\varepsilon$ some $\delta >0$ such that $U_1(x,\delta) \subseteq U_2(x,\varepsilon)$. Apply the definition of convergence under $d_1$ to $\delta>0$ and the $N$ found will also work for $d_2$ and $\varepsilon$, by the inclusion. etc.

Henno Brandsma
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