# Proving limit of bounded convergent sequence is bounded

Question: Show that if $$a \leq x_n \leq b$$ for every $$n$$ and $$x_n \rightarrow x$$, then $$a \leq x \leq b$$.

Proof: Let $$\epsilon>0$$. By assumption $$a_n \leq x_n \leq b$$ for all $$n$$. By definition of convergence, we have that there exists an $$N_1$$ such that $$|x_n-x|<\epsilon$$ for $$n \geq N_1$$.

Thus, when $$n \geq N_1$$, we have that $$x-\epsilon < x_n \leq b$$ and $$a \leq x_n < x+ \epsilon$$ Since $$\epsilon$$ is arbitrary, we have: $$x < b$$ and $$a Thus, $$a. Is my proof close to being correct, or am I not on the right track. Some confusions I'm having.

1. Can I say that $$\epsilon$$ being arbitrary implies that $$x and $$a.? Or is this an inaccurate statement? I have seen this argument in other proofs but don't fully understand the implication.
2. I proved that $$a but not $$a \leq x \leq b$$. How do I prove it for $$\leq$$ and $$\geq$$? Thanks and sorry for the seemingly trivial and basic questions.

Note that from $$|x_n-x|<\epsilon$$ you can write $$xtherefore $$a-\epsilonSince this holds for any $$\epsilon >0$$, we obtain $$a\le x\le b$$ This is where the $$\le$$ kicks in.
• Hi @Mostafa Ayaz isn't this just the same exact thing I showed? I don't see the difference, could you let me know the difference and where I messed up. But doesn't $a- \epsilon< x<b+\epsilon$ holding for any $\epsilon >0$ imply $a<x<b$? i don't see how it implies $a \leq x \leq b$. thank you – kemb Aug 24 '19 at 22:39
• Yes. This is almost exactly what you proved, except the last conclusion. For example let $$1-{1\over n}<x<2+{1\over n}$$ hold for any positive integer $n$. Does this mean $$x\in[1,2]$$ or $x\in (1,2)$? i.e. does $x=1$ or $x=2$ satisfy the inequality? – Mostafa Ayaz Aug 24 '19 at 22:49
• Hi @Mostafa Ayaz. thats what I don't understand. I don't understand why $1- \frac{1}{n}<x< 2+\frac{1}{n}$ holding for any positive integer n implies that $x \in [1,2]$. I thought it implied that $x \in (1,2)$. Could you explain why it implies that $x \in [1,2]$ as I don't understand this concept. Sorry for the basic questions – kemb Aug 24 '19 at 22:56
• It is OK. Do you agree that any $x_0$ for which the inequality holds for $n\in \Bbb N$, falls in the limit interval? – Mostafa Ayaz Aug 24 '19 at 22:57