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

Question

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

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 09 '15 at 03:31
@TravisJ Can you show how do i convert this to math language of proofs — Taylor Ted, May 09 '15 at 03:32
@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 09 '15 at 03:42
You are welcome. That we want something like $(x-b)/2$, or at least positive and less than $x-b$, is clear from the geometry, to make sure that if we back up by $\epsilon$ from $x$, we are still bigger than $b$. We also need a separate argument to show that $x$ is not less than $a$. — André Nicolas, May 09 '15 at 03:53
You can also use results from here for the sequence $(x_n)$ and two constant sequences. — Martin Sleziak, May 09 '15 at 05:28

score 3 · Accepted Answer · answered May 09 '15 at 03:35

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 $x_n\geq a\forall n$

score 3 · Answer 2 · answered May 09 '15 at 03:40

3

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.

answered May 09 '15 at 03:40

TravisJ

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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 09 '15 at 04:03
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@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 09 '15 at 04:08

HorizonsMaths · Answer 3 · 2015-05-09T06:04:28.350

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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 $a\le x\le b$.

edited May 09 '15 at 06:04

answered May 09 '15 at 05:07

HorizonsMaths

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

3 Answers3