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Question:

Let $N=N_1\cup...\cup N_k$ . If $n \in N_p$, denote the subsequence $x_n := x^p_n$ for $p=1,...k$. Suppose that $\lim x^1_n=...=\lim x^k_n=a$. Prove that $\lim x_n = a$, for $n\in N$.

My attempt:

EDIT:

As David noticed, it is complete non-sense (and unjustified) to write $x_n = \sum_{p=1}^{k} x^p_n$. So i changed the proof. Here it is:

Since $\lim x^1_n=...=\lim x^k_n=a$, there exists $n_1,...,n_k$, with $n_1 \in N_1,..., n_k \in N_k$, such that:

for $n\in N_1$, $ n>n_1$ implies that $ x_n^1 \in(a-\epsilon, a+ \epsilon), ....,$ and $n \in N_k, n> n_k$ implies that $x_n^k\in(a-\epsilon, a+ \epsilon)$.

Taking $n_o=$max$\{n_1,...n_k\}$, for $n \in N_1\cup...\cup N_k$ and $n>n_o$ it follows that $ x_n^1,...,x_n^k \in(a-\epsilon, a+ \epsilon)$. Since $N=N_1\cup...\cup N_k$ by hyphotesis, then $n \in N$.

Therefore, $n\in N$ and $n>n_o$ imples that $x_n \in(a-\epsilon, a+ \epsilon)$.

Hence, $\lim x_n = a.$

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    Why is $x_n = \sum_{p=1}^{k} x^p_n$? – 2017-01-18
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    Why and where do you use that $k$ is even? Also, *can* you assume it wlog? – 2017-01-18
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    Since $k$ is natural, it is even or odd. That's why i assumed wlog that it was even. I used the fact that it was even to write down the sum $a-a+...-a+a = 0$, using summation notation. – 2017-01-18
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    I changed the title because the original title doesn't correspond to the question in this problem, unless you want to say that the hypothesis hold for every $k$ and every $N_1,...,N_k$ in which case the proof is one sentence. – 2017-01-18
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    Yes, it is unjustified and non-sense to write down $x_n$ as a sum of its subsequences. It sounded intuitive (at least while i was writing the proof), but it is absolute non-sense. I've changed the proof, since the last one is completly wrong. – 2017-01-18

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I am skeptical of this proof. It could definitely be better explained (I find it hard to follow) but from what I see I don't see any reason to be able to way "WLOG $k$ is even." Why is that true?