Proving a sequence is convergent using subsequences.

Question

I recently learned how to prove that showing $(x_n)$ is converges to $L$, it is sufficient to show that it's subsequence of even and odd terms converge to $L$

But I wonder, can we do better and generalise a bit? I'm going to write down some of my thoughts, I might lack some rigour so please excuse me.

Let $(x_n)_{n \in \mathbb{N}}$ be a real sequence. Let $T_1, T_2,\ldots, T_k \subseteq \mathbb{N}$ be infinite subsets of $\mathbb{N}$ such that $\bigcup_{i=1}^k T_i = \mathbb{N}$ then, can I claim something like:

If $(a_n)_{n \in T_i}$ converges to $L$ for each $i$ then $\lim\limits_{n \to ∞} a_n =L$

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My intuition tells me this should be true although I'm not sure how I would begin to prove such a thing. If it's not true, then... well why?

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Proof 1:

If $(a_n)_{n \in T_i} \to L, \forall \, i \, (1≤i≤k)$ then $$\forall \, \epsilon >0, \exists \, N_i \in \mathbb{N}, N_i \in T_i : \forall \, n ≥N_i, n \in T_i \Rightarrow |a_n-L| < \epsilon $$ Let $N = \max\limits_{1≤i≤k} N_i$ then $$\forall \, n ≥N, n \in T_i \Rightarrow |a_n-L| < \epsilon, \forall \, i \, (1≤i≤k)$$ i.e. $ \forall \ n \, (n \in T_1$ or $n \in T_2$ or $\ldots$ or $n \in T_k, n ≥N) \Rightarrow |a_n-L| < \epsilon$ and thus, $$\forall \, n ≥N, n \in \bigcup_{i=1}^n T_i \Rightarrow |a_n-L| < \epsilon $$ or $$\forall \, n ≥N, n \in \mathbb{N} \Rightarrow |a_n-L| < \epsilon$$ Hence, the proof is complete. Is this correct?

Proof 2: This proof will be rewriting proof 1 but with alternative notation which is the subsequence notation.

Let $T_i = \{t_n^i \in \mathbb{N}: n \in \mathbb{N} \}\subseteq \mathbb{N}$, note that $t_n^i$ denotes the $n$th position in the sequence of $T_i$ i.e. $(a_n)_{n \in T_i} = (a_{t_{n}^i} )_{n \in \mathbb{N}}$ as $(t_n^i)$ is a strictly increasing sequence in $T_i$.

Such a sequence $(t_n^i)$ exists in $T_i$ because $T_i$ is an infinite subset of $\mathbb{N}$ with the usual ordering of $\mathbb{N}$ so I can construct a strictly monotone sequence in $T_i$ like: $t_1^i <t_2^i <t_3^i< \ldots<t_n^i<\ldots $ . I'm not sure how to write this part more rigourously.

The rest of the proof would be similar to 1 by simply choosing a maximum $N_i$.

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Answer 1

The proof in your question is probably correct. I find it almost undreadable, and so have not checked it. Here is how I would write it, using words as much as possible.

Given an $\epsilon > 0$ I want to find an integer $N$ beyond which all the sequence elements $a_n$ are within $\epsilon$ of $L$. (That's just using the definition of the limit.)

Well for each of the finitely many subsequences indexed by a $T_i$ I can find an $N_i$ that does the job for that subsequence. Now let $N$ be the maximum of the $N_i$. Then for every $n>N$ I know $a_n$ is in one of the $i$ subsequences since the union of the $T_i$ is all of $\mathbb{N}$. Then $a_n$ is within $\epsilon$ of $L$.

I suspect that is pretty much what you were thinking when you wrote your proof. There's no point in rewriting clear thought in English as strings of symbols.

Note where the finiteness is used. That suggests that the claim might fail if you don't assume that there are just finitely many subsequences. So you should either look for a counterexample or try to write the proof without the assumption.

Edit in response to comments.

Is the assertion true if you say

Let $T_1,T_2, \ldots ⊆ \mathbb{N}$ be infinitely many subsets of $\mathbb{N}$ whose union is $\mathbb{N}$ then ...

Excessive (that is, unnecessary) use of symbols rather than words makes what you write hard to read. It's definitely unprofessional. See Is it bad form to write mysterious proofs without explaining what one intends to do?