Completeness of the union

Question

I want to prove if $M_1, M_2 \subset N$ are complete, then $M_1 \cup M_2$ is complete.

I was trying taking an arbitrary Cauchy sequence in $M_1 \cup M_2$, and trying to find a convergent subsequence in $M_1$ or $M_2$: If $(x_n)\subset M_1 \cup M_2$ is a Cauchy sequence, wlog there exists $(x_{n_k})\subset M_1$, this is a Cauchy subsequence. (Is this right?). But $M_1$ is complete, then $(x_{n_k})\to x \in M_1$. Then $(x_n)\to x \in M_1 \subset M_1 \cup M_2$. As $(x_n)$ was an arbitrary Cauchy sequence, we can conclude that $M_1 \cup M_2$ is complete.

Someone can help? Is the first time I try to do this kind of problems.

Answer 1

Assumming N is not the natural numbers, which would not be complete.

Let $\{p_n\}$ be a Cauchy sequence in one of the set M$_1$ or M$_2$. Choose M$_1$ without loss of generality. The $\{p_n\}\subset M_1$. Since M$_1$ is complete, $\{p_n\}$ converges to a point of M$_1$. M$_1\subset M_1\cup M_2$. Thus $\{p_n\}$ converges to a point of $M_1\cup M_2$.

Answer 2

Given a Cauchy sequence in $M_1\cup M_2$, it must have infinitely many (not necessarily distinct) terms in $M_1$ or infinitely many in $M_2$, because if it had only finitely many terms in each then it would have only finitely many terms altogether. So suppose it has infinitely many terms in $M_i$. Those terms constitute a Cauchy sequence, because they're a subsequence of the original Cauchy sequence. So this subsequence converges to some point in $M_i$, because $M_i$ is complete. But then the whole original sequence converges to the same point, because a Cauchy sequence with a convergent subsequence is always convergent (regardless of any completeness assumptions).