# If $(x_{n})$ is sequence with $x_{n} \leq b$ for all $n$, then $\text{lim}_{n \to \infty} x_{n} \leq b$

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

• "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 '18 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 '18 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 '18 at 15:50
• I think I only needed to assume that the limit exists for my proof to work – Sei Sakata Dec 6 '18 at 15:52
• 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 $\neg (\lim_{n\to \infty} x_n=c).$ – DanielWainfleet Jan 31 at 3:14