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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.

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  • $\begingroup$ Shouldn't the interval $[c,b]$ be actually $[b,c]$? $\endgroup$ – Bernard Oct 20 '18 at 16:25
  • $\begingroup$ @Bernard that doesn't matter in this sense. $c$ and $b$ are just arbitrary numbers. $\endgroup$ – dc3rd Oct 20 '18 at 16:33
  • $\begingroup$ I'm sorry, but as you wrote it, there's no contradiction in having $a>c$, whereas there is one with $[b,c]$. $\endgroup$ – Bernard Oct 20 '18 at 16:50
  • $\begingroup$ @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... $\endgroup$ – dc3rd Oct 20 '18 at 16:53
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    $\begingroup$ "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. $\endgroup$ – fleablood Oct 20 '18 at 18:02
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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}<a_n < 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}<a_n$ for all $ \ n \geq N$.

Do you see the contradiction now?

Can you do the same for $a<c$?

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    $\begingroup$ 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 $N$. $\endgroup$ – John11 Oct 20 '18 at 16:55
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    $\begingroup$ @dc3rd Yes I corrected it. $\endgroup$ – John11 Oct 20 '18 at 16:58
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    $\begingroup$ @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. $\endgroup$ – John11 Oct 20 '18 at 17:08
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    $\begingroup$ John11 has said better what i was saying last night (before I went to sleep). $\endgroup$ – Randall Oct 20 '18 at 17:10
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    $\begingroup$ 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. $\endgroup$ – dc3rd Oct 20 '18 at 17:12
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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 $\beta<c.$ 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 $-\frac{\beta -\alpha }{2}<a-a_n<\frac{\beta -\alpha }{2}$ 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.

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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 $a \in [c,b]$.

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