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Given that $X$ is a dense subset of the metric space $Y$ and that all Cauchy sequences in $X$ are convergent to points in $Y$ I want to prove that $Y$ is complete.

Here's my attempt so far:

Let $\{y_n\}$ is a Cauchy sequence in $Y$. Assume that $\{y_n\}$ has a subsequence $\{x_n\} \in X$. Then $\{x_n\}$ is also Cauchy (acc. to theorem), and we know that Cauchy sequences in $X$ are convergent. As a subsequence of $\{y_n\}$ is convergent, so is $\{y_n\}$ with the same limit point (acc. to theorem).

Now assume $\{y_n\}$ is a Cauchy sequence with only a finite number of elements within $X$, and infinitely many on $Y-X$. I want to prove that $\{y_n\}$ is convergent also in this case, but I am not sure how to proceed. I have tried to work with a subsequence here too in which I exclude all elements in $X$. That is, I've tried to prove $\{y_{n_k}\} = \{y_i|y_i\not\in X\}$ is convergent, but failed.

Any ideas on how the proof can be finished?

1 Answers 1

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You have not made use of the hypothesis that $X$ is dense.

Let me give just a hint:

Write your sequence as $y_1,y_2,y_3,...$ and ponder each term one at a time.

Knowing that $X$ is dense, what can you say about $y_1$?

Knowing that $X$ is dense, what can you say about $y_2$?

And so on and so on...

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    $(x_i,k)_{k \in \Bbb N} \to y_i \forall i \in \Bbb N$. But how can we get a Cauchy sequence in $X$ from this? I think that's where the OP gets stuck. – 2017-02-18
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    It looks to me as if the OP is getting stuck on the thought that some points of $y_i$ are in $X$ and some aren't, which is not the correct focus for this problem. I am attempting to change the focus, namely how the hypothesis density of $X$ is to be applied. – 2017-02-18