Prove: $x_{n}\rightarrow a\iff d(x_{n},a)\rightarrow 0$

Question

Prove: $$x_{n}\rightarrow a\iff d(x_{n},a)\rightarrow 0$$

Intuitively and in euclidian metric it seems to be trivial, but I am stuck.

$(\Rightarrow):$ There for for all $\epsilon>0$ there is $n>N$ such that $d(x_{n},a)<\epsilon$

And I need to get to $d(d(x_{n},a),0)$

score 2 · Answer 1 · answered Apr 27 '17 at 13:51

2

In $d(d(x_n,a),0)$ two $d$ are not the same metric. The last one is just $|d(x_n,a)|$ as is in $\mathbb{R}$.

answered Apr 27 '17 at 13:51

SaeidAli

144

This is a comment to the question rather than an answer. – robjohn Apr 27 '17 at 13:55
This is not a comment, this is the answer: as @gbox asked: $d(d(x_n,a),0)$ is required. my answer is $d(d(x_n,a),0)=|d(x_n,a)|=d(x_n,a)<\varepsilon$. – SaeidAli Apr 27 '17 at 13:59
1

Okay, I see where you're going, but I think a bit more clarification might be useful. (+1) – robjohn Apr 27 '17 at 14:51
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I agree. I also needed to put that a metric on $X$ is $d:X\times X\rightarrow\mathbb{R}$. – SaeidAli Apr 27 '17 at 14:55

score 2 · Accepted Answer · answered Apr 27 '17 at 14:49

2

If $d$ is a metric, then $d(x_n,a)\ge0$. Therefore, you are given $$ \text{for all }\epsilon\gt0\text{, there is an }N\text{ so that if }n\ge N\text{, then }0\le d(x_n,a)\lt\epsilon $$ Thus, for $n\ge N$, $$ d(d(x_n,a),0)=|d(x_n,a)-0|\lt\epsilon $$ The outer $d$ is in $\mathbb{R}$ while the inner $d$ is in the other metric space.

answered Apr 27 '17 at 14:49

robjohn

345,667

Why can we say that the outer d is in $\mathbb{R}$? – gbox Apr 27 '17 at 14:52
1

Because $d(x_n,a)\in\mathbb{R}$ and $0\in\mathbb{R}$, so the outer $d$ is measuring distance in $\mathbb{R}$. – robjohn Apr 27 '17 at 14:54
1

A metric over a set $X$ is defined as : $$d:X\times X\rightarrow\mathbb{R}$$ what ever your metric is. – SaeidAli Apr 27 '17 at 14:57

Prove: $x_{n}\rightarrow a\iff d(x_{n},a)\rightarrow 0$

2 Answers2