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Let $X$ be a separable metric space and $A \subset$ X be countable and dense. Characterize the statements below as true or false (and why).

  • If every Cauchy sequence in $A$ converges in $X$, $A$ is complete.

  • If every Cauchy sequence in $A$ converges in $X$, $X$ is complete.

  • Is $X = \mathbb{R} \setminus \{\sqrt{2}\}$ and $A = \mathbb{Q}$ a direct counterexample of the two above statements?

For the first statement, I would say that it's false, since every Cauchy sequence of $A$ does not converge in $A$, so $A$ is not necessarily complete.

As for the second, there may be divergent Cauchy sequences of $X$, so false again.

Is my approach correct and if so, does the third statement play the role of a counterexample of the aforementioned?

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Your approach is not correct.

The first statement is “If every Cauchy sequence in $A$ converges in $X$, $A$ is complete.” You cannot deduce from this that not every Cauchy sequence of $A$ converges in $A$. What If it turns out that $X=A$?

The second stament is “If every Cauchy sequence in $A$ converges in $X$, $X$ is complete.” Concerning this, what you do is to claim that there may be divergent Cauchy sequences of $X$. That's wrong. There is no such sequence.

And $X=\mathbb R\setminus\left\{\sqrt2\right\}$ and $A=\mathbb Q$ is a counterexample to the first statment, but not to the second one: $\left(\frac{\left\lfloor\sqrt2n\right\rfloor}n\right)_{n\in\mathbb N}$ is a Cauchy sequence of elements of $\mathbb Q$, but it does not converge in $\mathbb R\setminus\left\{\sqrt2\right\}$.

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For the first statement $A=\mathbb Q$, $X=\mathbb R$ gives a counterexample.

The second statement is true. Proof: let $\{x_n\}$ be a Cauchy sequence in $X$. For each $n$ there exists $y_n \in A$ such that $d(x_n,y_n) <\frac 1 n$. Now it is easy to verify (by triangle inequality) that $\{y_n\}$ is a Cauchy sequence in $A$. By assumption $y_n$ converges to some $y$ and triangle in equality helps you again to show that $x_n \to y$. Hence $X$ is complete. Obviously the third statement is false.

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