If $(x_{n})$ as sequence with $x_{n} \leq b$ for all $n$, then $\text{lim}_{n \to \infty} x_{n} \leq b $ if limit exists.

I see this fact being used very often and it seems fairly trivial but I am wondering how it can be proven rigorously.

Proof attempt:

Suppose that the limit exists.

Suppose $x_{n}<b$ for all $n \in \mathbb{N}$. Then by definition of limit, for any arbitrary $\epsilon>0$, we can find $N \in \mathbb{N}$ such that for every $n^*>N$ we have $|(\text{lim}_{n \to \infty} x_{n})-x_{n^*}|< \epsilon$. Then $\text{lim}_{n \to \infty} x_{n}<\epsilon +x_{n^*} \leq \epsilon +b$

So $\text{lim}_{n \to \infty} x_{n}< \epsilon + b$ where $\epsilon$ is arbitrary.

This implies that at most, $\text{lim}_{n \to \infty} x_{n} \leq b$

  • "Assume $x_n$ is increasing and convergent without loss of generality" : how are you sure that you are not losing generality here? – астон вілла олоф мэллбэрг Dec 6 at 15:45
  • Maybe it's not obvious if it's not monotonic but we can construct a monotonic subsequence which has the same limit. But in the case of decreasing limit, I think its pretty obvious? – Sei Sakata Dec 6 at 15:48
  • You do not seem to have used the fact that it is increasing in the proof anyway. The proof is fine. Also, more is true : even if $x_n$ is not convergent, the limit superior of $x_n$ is smaller than $b$. – астон вілла олоф мэллбэрг Dec 6 at 15:50
  • I think I only needed to assume that the limit exists for my proof to work – Sei Sakata Dec 6 at 15:52
  • Yes. In that respect, this proof is fine. You proof is rigorous enough for my liking. – астон вілла олоф мэллбэрг Dec 6 at 15:53

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