Union of Sequences My Analysis professor mentioned the following theorem: If a sequence $(a_n)$ is the union of finitely many disjoint subsequences, and if all the subsequences converge to $l$,then sequence $(a_n)$ converges to $l$.
I'm not entirely sure whether this language is appropriate. Could someone correct it/help me understand? 
 A: Sequences are functions which in turn are sets and union of sets is nothing weird. The union of functions need not be a function. A sufficient, but not necessary condition for the union of functions to be a function is that the domains are disjoint.
As the OP has pointed out in this comment, the union of disjoint subsequences of a given sequence $a$ need not be a sequence. Below is an example of this.
Recall that a subsequence of $a$ is a sequence of the form $a\circ \alpha$, where $\alpha$ is strictly increasing sequence of natural numbers.
Let $a=\text{id}_{\mathbb N}, \alpha=\{(n,2n)\colon n\in \mathbb N\}$ and $\beta=\{(n,2n-1)\colon n\in \mathbb N\}$. One gets $a\circ \alpha=\alpha, a\circ \beta=\beta$ and clearly $\alpha\cup \beta$ is not a function despite $\alpha \cap \beta=\varnothing$, that is, the union of the subsequences $a\circ \alpha$ and $a\circ \beta$ is not a sequence.
What the author of the problem should have said is "if $a$ has the property that finitely many of its subsequences $a\circ\alpha _1, \ldots ,a\circ \alpha _n$ are such that the range of any pair of $\alpha _k$'s is disjoint and the union of all the ranges of $\alpha_k$'s is $\mathbb N$ (which is the domain) of $a$, then ..."
