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I am trying to prove that if $\{a_n\}$ is an increasing (non-decreasing) sequence with $a_n \to a$, then $\{a_n\}$ is bounded above and $a = \sup a_n$.

Obviously, I understand this to be true intuitively, but am struggling with a rigorous proof. I have started by declaring an $\epsilon>0$ and saying that given a natural number $N$, $n \ge N$, that $\vert a_n-a \vert < \epsilon$. I'm having difficulty actually proving that the sequence is bounded and that a is in fact the supremum.

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You need to show

  • For all $n$, $a_n\le a$
  • For all $b<a$, there exists $n$ with $a_n>b$.

For the first part, suppose that $a_n>a$ for some $n$. By convergence, almost all $a_k$ differ from $a$ by less than $\epsilon := a_n-a$. In particular there exist $k>n$ with $a_k<a+\epsilon = a_n$, contradicting the non-increasing condition.

For the second part: If $b<a$, then almost all $a_n$ differ from $a$ by less than $\epsilon:=a-b$, i.e., almost all $a_n$ are $>b$.

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  • $\begingroup$ thanks. but i still don't see how we're saying all a must be less than or equal to a. Because I can't say that any term minus a EQUALS epsilon, just that it must be less than epsilon. so i dont think i can use that equality statement. $\endgroup$ – mayalarson Feb 10 at 21:18
  • $\begingroup$ The condition needs to hold for all $\varepsilon > 0$. Assuming $a_n > a$, we have that $a_n-a >0$, why this positive real number can be used to check the condition as an "epsilon" @mayalarson $\endgroup$ – ms_ Feb 11 at 17:51
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let $a=sup\{a_n:n\in \mathbf{N}\}$ now since a is a least upper bound of $a_n$ if $\epsilon >0$ then $a-\epsilon$ is no longer an upper bound so $\exists N\in \mathbf{N}$ so than $a-\epsilon <a_N<a$ but $a_n$ is increasing so $\forall n>N $ we have $a_N<a_n<a$ and hence $|a_n-a|<\epsilon$ so $a_n\rightarrow a$

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