# if $\{a_n\}$ is a sequence such that $a_n \in [c,b]$ and $a_n \rightarrow a$ then $a \in [c,b]$ not understanding the contradiction

if $$\{a_n\}$$ is a sequence such that $$a_n \in [c,b]$$ and $$a_n \rightarrow a$$ then $$a \in [c,b]$$

I just wanted to make sure I was writing a correct proof and communicating my ideas correctly.

Proof

Suppose $$a \notin [c,b]$$. Given that $$a_n \rightarrow a$$ this means: $$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon$$

$$\Rightarrow \ \exists \ a_n \ s.t. \ a_n \notin [c,b]$$

Not sure if I have to give this explanation after, but since $$a_n$$ converges to $$a$$ it means eventually that $$a_n$$ and $$a$$ are going to be so close or equal that $$a_n$$ will not be able to be in $$[c,b]$$

EDIT 2 - Revised with correct solution

I'm going to leave up my original solution as well so others can possibly see the difference between what would be considered a sound proof VS one that is not even if you may have the right idea.

Proof

Let $$\epsilon >0$$ and let $$a < c \Leftrightarrow 0 < c-a$$.

Given that $$a_n \rightarrow a$$ this means: $$\forall \ \epsilon >0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N \ |a_n - a| < \epsilon$$

Since the claim of convergence has to work for all $$\epsilon$$, consider $$\epsilon = \frac{c-a}{2}$$

Since $$a_n \rightarrow a$$:

$$|a_n - a| < \epsilon = \frac{c-a}{2}\\ \Rightarrow \\ \frac{-(c-a)}{2} + a < a_n < \frac{c-a}{2} + a = \frac{c}{2} + \frac{a}{2}$$

We assumed $$a < c$$

$$\Rightarrow \frac{c}{2} + \frac{a}{2} < \frac{c}{2} + \frac{c}{2} = c$$ $$\therefore a_n < \frac{c}{2} + \frac{a}{2} < c \\ \Rightarrow \ a_n \notin [c,b] \ \forall \ n \geq N$$

But one of our assumptions is that $$a_n \in [c,b]$$. So we have a contradiction.

• Shouldn't the interval $[c,b]$ be actually $[b,c]$? – Bernard Oct 20 '18 at 16:25
• @Bernard that doesn't matter in this sense. $c$ and $b$ are just arbitrary numbers. – dc3rd Oct 20 '18 at 16:33
• I'm sorry, but as you wrote it, there's no contradiction in having $a>c$, whereas there is one with $[b,c]$. – Bernard Oct 20 '18 at 16:50
• @Bernard ok, going with [b,c] as the interval. I suppose the contradiction would be the one I'm trying to see in the discussion below? ....which for the life of me I am still not finding... – dc3rd Oct 20 '18 at 16:53
• "Not sure if I have to give this explanation after, but since an converges to a it means eventually that an and a are going to be so close or equal that an will not be able to be in [c,b]" I think this is skirting around the idea that if $a \not \in [c,b]$ then $\inf \{|a-k|: k \in [c,b]\} > 0$ and so when $|a_n - a| < \epsilon = \inf \{|a-k|: k \in [c,b]\}$ we can conclude $a_n \not \in [c,b]$. .... This is acceptable but you must demonstrate that if if $a \not \in [c,b]$ then $\inf \{|a-k|: k \in [c,b]\} > 0$. Which is not a given. – fleablood Oct 20 '18 at 18:02

Suppose $$a>b$$. Then for $$\epsilon =\frac{a-b}{2}>0,\ \exists \ N\in \mathbb{N} \ s.t.\ \forall \ n \geq N$$ we have $$\ |a_n - a| < \frac{a-b}{2}$$.

But then this means: $$a-\frac{a-b}{2}

Looking at the left-hand side we see: $$b=\frac{b+b}{2}<\frac{b+a}{2}=a-\frac{a-b}{2} for all $$\ n \geq N$$.

Do you see the contradiction now?

Can you do the same for $$a?

• No. It is that we supposed $a_n$'s are always in the interval $[c,b]$. So they have to be smaller than $b$ but at the end we find $a_n>b$ starting a certain rank $N$. – John11 Oct 20 '18 at 16:55
• @dc3rd Yes I corrected it. – John11 Oct 20 '18 at 16:58
• @dc3rd Yes. Basically you describe the idea of the proof without explicitly explaining a way to find such an $a_n$ that is not in the interval. That is not acceptable. You don't convince the reader you understand why such $a_n$'s would exist. Taking $(a-b)/2$ for $\epsilon$ works but so would have $(a-b)/4$ etc. You need to show you understand the meaning of definition and actually use it. – John11 Oct 20 '18 at 17:08
• John11 has said better what i was saying last night (before I went to sleep). – Randall Oct 20 '18 at 17:10
• Sorry for not completely getting what you said @Randall, but introducing the half distance trick definitely helped. Since it has to work for all epsilon, if I choose an "arbitrary" one then it better work for that one or like in this scenario I get a contradiction. – dc3rd Oct 20 '18 at 17:12

You are on the right track. If $$a\notin [c,b]$$ then $$a\in [c,b]^c,$$ which is an open set in $$\mathbb R.$$ So, $$a$$ is either contained in some open interval to the left of $$c$$ or to the right of $$b$$. Drawing a picture will help.

Let's say $$a$$ is in the center of an interval $$(\alpha,\beta)$$ with $$\beta Now, since $$a_n\to a$$, it must be the case that, if $$n$$ is large enough, all $$a_n$$ land in $$(\alpha, \beta).$$ More precisely: let $$\epsilon =\frac{\beta -\alpha }{2}$$. Then, there is an integer $$N$$ such that if $$n>N$$, $$|a-a_n|<\epsilon=\frac{\beta -\alpha }{2}.$$ This means that $$-\frac{\beta -\alpha }{2} so the distance between $$a$$ and $$a_n$$ is less than half the width of the interval $$(\alpha,\beta).$$ That is to say, the $$a_n$$ are in $$(\alpha, \beta),\$$ which is a contradiction because $$(\alpha, \beta)$$ is disjoint from $$[c,b]$$ by construction.

Perhaps instead of proving that $$a \in [c,b]$$, prove that if $$k \not \in [c,b]$$ then $$k \ne a$$.

If $$k < c$$ and $$a_n \in [c,b]$$ then $$a_n \ge c > k$$ and $$|a_n - k| = a_n - k = (a_n -c) + (c-k) \ge c-k > 0$$. So for $$\epsilon = c-k$$ then $$|a_n - k| \ge \epsilon$$ for all $$a_n$$. So $$a_n \not \to k$$.

Likewise if $$k > b$$ and $$a_n \in [c,b]$$ then $$k > b \ge a_n$$ and $$|a_n - k| = k- a_n = (k-b) + (b-a_n) \ge k-b > 0$$. So for $$\epsilon = k - b$$ then $$|a_n - k| \ge \epsilon$$ for all $$a_n$$. So $$a_n \not \to k$$.

So if $$k \not \in [c,b]$$ then $$a_n \not \to k$$.

So if $$a_n \to a$$ by contrapositive $$a \in [c,b]$$.