Let $X$ and two strongly equivalent metrics $d_1, d_2 : X \times X \to \mathbb R$ on $X$ be given. Prove that a sequence $x_n$ converges to an element $x_0$ of $X$ in $(X, d_1)$ if and only if the sequence converges to $x_0$ in $(X, d_2)$.
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1I was being a bit flippant. But in general, it helps to say what it is you don't understand. Are you uncertain about what the question is asking? Do you know what you need to show, but are unsure about how to do so? – Math1000 Mar 11 '15 at 02:57
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I have a general idea of what it is asking, but honestly I'm kind of lost. I understand that $d_1$ and $d_2$ differ by a constant? But I'm not sure how to use the definition of convergence in the proof. – Shannon Mar 11 '15 at 03:39
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Since $d_1$ and $d_2$ are strongly equivalent, there exist constants $c,C>0$ such that for all $x,y\in X$, $$cd_1(x,y) \leqslant d_2(x,y) \leqslant Cd_1(x,y). $$ Suppose $x_n\to x_0$ in the $d_1$ metric. Let $\varepsilon>0$. Choose $N$ so that $n\geqslant N$ implies $$d_1(x,x_n)<\frac\varepsilon C.$$ Then if $n\geqslant N$, $$d_2(x,x_n)\leqslant Cd_1(x,x_n)<C\cdot\frac\varepsilon C =\varepsilon, $$ so that $x_n\to x_0$ in the $d_2$ metric. The converse can be proved similarly.
Math1000
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HINT:the collection of open sets in two topologically equivalent metrics is same .Now use the definition for a sequence to be convergent
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