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If we have an increasing sequence of sets, $A_n \subset A_{n+1}$, prove that the limit of this sequence not only exists but is the union of the sets. i.e. $ A_n \uparrow\cup_{n=1}^{\infty}A_n$.

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How do you think you should start? What have you tried? What are your thoughts? – anon Sep 30 '12 at 4:10
The first thing to do is to make sure you know the definition of a limit of a sequence of sets. – Gerry Myerson Sep 30 '12 at 4:10

How is the limit (for sequence of sets) is defined?

Following the definition I studied, $\exists\lim A_i \iff \limsup A_i = \liminf A_i$, where $$\limsup_n A_n := \bigcap_{n=0}^\infty \bigcup_{k=n}^\infty A_k$$ $$\liminf_n A_n := \bigcup_{n=0}^\infty \bigcap_{k=n}^\infty A_k$$

Now, if $A_n$ is monotonic ($A_n\subseteq A_{n+1}$), try to evaluate these, and you will get their union in both cases.

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