Let $a \leq x_{n} \leq b$ for all n in N. If $x_{n} \rightarrow x$. Then prove that $a \leq x \leq b$

Let $a \leq x_{n} \leq b$ for all n in N. If $x_{n} \rightarrow x$. Then prove that $a \leq x \leq b$

Attempt - If I assume that $x$ is greater than both $a$ and $b$. Then since series is given convergent, so after certain stage its elements will lie between $(x-\epsilon , x + \epsilon )$. If I take epsilon to be such that $\epsilon = (b + x) /2$. Then sequence lies to right of $b$, which is contradiction. Same argument for if I take limit to be to left of a. Is this fine? Thanks

• Yes, your argument is great! :) – layman May 9 '15 at 3:31
• Looks good. Exactly the right idea. If you get close to a number to the right of $b$, eventually you must be to the right of $b$. – TravisJ May 9 '15 at 3:31
• @TravisJ Can you show how do i convert this to math language of proofs – Taylor Ted May 9 '15 at 3:32
• I think you want something like $\epsilon=(x-b)/2$. – André Nicolas May 9 '15 at 3:40
• @K.Dutta, I wrote an answer that (verbosely) spelled out how to make this formal. The take-home is make $\epsilon$ such that when $|x_n - x|<\epsilon$ then you know you are outside of the interval. – TravisJ May 9 '15 at 3:42

I am showing one side only.

If $x<a$ then taking $\epsilon=a-x>0$

Now since $x_n \to x \exists m\in \mathbb N$ such that $x-\epsilon<x_n<x+\epsilon \forall n\geq m$

Then $x_n<x+\epsilon=x+a-x=a$ i.e $x_n<a$ for some $n$ contradiction as $x_n\geq a\forall n$

You are on the right track. As requested in the comment, the formalism is as follows:

If $x>b$, take $\epsilon=x-b$. Then if $n>N$ it follows that

$$|x_{n}-x| < x-b.$$

We know that this should mean that $x_n>b$ (which is a contradiction) so let's re-write the terms to make this appear very explicitly.

$$-(x-b) < x_{n}-x < (x-b)$$

Adding $x$ to all three parts of the inequality gives us:

$$b < x_{n} < 2x-b$$

but of course, we only need that $b<x_{n}$ to get the contradiction.

• Sorry for not accepting the your answer Only reason was that other user posted before you .So according to ethics i accepted that ..Your answer is fantastic though – Taylor Ted May 9 '15 at 4:03
• @K.Dutta, no worries. But, to be clear, the answer you accept is the one you feel best answers your question. I'm never offended when people don't select my answer--feel free to select as you see fit. – TravisJ May 9 '15 at 4:08

Given $\varepsilon>0$, there is a positive integer $N$ such that $$|x-x_n|\le \varepsilon \quad \forall n\ge N.$$ It follows that $$x_n-\varepsilon \le x\le x_n+\varepsilon \quad \forall n\ge N$$ and therefore $$a-\varepsilon \le x\le b+\varepsilon.$$ Letting $\varepsilon$ tend to $0$, we get $a\le x\le b$.