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 $x_{n}\leq b$ 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, $\text{lim}_{n \to \infty} x_{n} \leq b$

  • $\begingroup$ "Assume $x_n$ is increasing and convergent without loss of generality" : how are you sure that you are not losing generality here? $\endgroup$ – астон вілла олоф мэллбэрг Dec 6 '18 at 15:45
  • $\begingroup$ 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? $\endgroup$ – Sei Sakata Dec 6 '18 at 15:48
  • $\begingroup$ 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 $b$. $\endgroup$ – астон вілла олоф мэллбэрг Dec 6 '18 at 15:50
  • $\begingroup$ I think I only needed to assume that the limit exists for my proof to work $\endgroup$ – Sei Sakata Dec 6 '18 at 15:52
  • 1
    $\begingroup$ Good proof. ..BTW $\lim_{n\to \infty}x_n=x$ is equivalent to: "For all $r>0$ the set $\{n\in \Bbb N: x_n\not \in (x-r,x+r)\}$ is finite." So if every $x_n\le b$ then for any $c>b$, consider $ r_c=(c-b)/2>0.$ Then the set $\{n\in \Bbb N: x_n\not \in (c-r_c,c+r_c)\}$ not only fails to be finite; it is all of $\Bbb N,$ so $\neg (\lim_{n\to \infty} x_n=c).$ $\endgroup$ – DanielWainfleet Jan 31 at 3:14

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