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 Post subject: Metric space and dense subsetPosted: Wed Mar 09, 2016 6:12 pm
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Let $\displaystyle{\left(X,d\right)}$ be a metric space and $\displaystyle{D}$ a dense subset of $\displaystyle{X}$ having the

property : Each $\displaystyle{\rm{Cauchy}}$ sequence of elements of $\displaystyle{D}$ converges to $\displaystyle{X}$ .

Prove that the metric space $\displaystyle{\left(X,d\right)}$ is complete.

Comment : The metric space $\displaystyle{\left(\mathbb{R},|\cdot|\right)}$ with $\displaystyle{D=\mathbb{Q}}$ is an example of such space.

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 Post subject: Re: Metric space and dense subsetPosted: Wed Mar 09, 2016 6:13 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 838
Location: Larisa
Papapetros Vaggelis wrote:
Let $\displaystyle{\left(X,d\right)}$ be a metric space and $\displaystyle{D}$ a dense subset of $\displaystyle{X}$ having the

property : Each $\displaystyle{\rm{Cauchy}}$ sequence of elements of $\displaystyle{D}$ converges to $\displaystyle{X}$ .

Prove that the metric space $\displaystyle{\left(X,d\right)}$ is complete.

Comment : The metric space $\displaystyle{\left(\mathbb{R},|\cdot|\right)}$ with $\displaystyle{D=\mathbb{Q}}$ is an example of such space.

Pick a Cauchy sequence say $x_n$. Since the set $\displaystyle{D}$ is dense there exist for every $n$ an element $y_n \in D$ such that $d(x_n, y_n)<\frac{1}{n}$. Using the relation:

$$d(y_m, y_n)\leq d(y_m, x_m) + d(x_m, x_n) + d(x_n, y_n)$$

we can easily see that $y_n$ is Cauchy. Hence converges. Let $x \in X$ be its limit. Since:

$$0\leq d(x_n, y) \leq d(x_n, y_n) +d(y_n, y)$$

we easily see that $x_n$ converges to $y$, completing the proof.

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