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

Question

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.

Answer 1

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 $a<c$?

Answer 2

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.

Answer 3

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 $a \in [c,b]$.