If $\left\{x_n\right\}$ is a convergent sequence of points in $[a, b]$ and $\lim x_n = c$, then $c\in[a, b]$

Question

If $\left\{x_n\right\}$ is a convergent sequence of points in $[a, b]$ and $\lim x_n = c$, then $c\in[a, b]$.

This is a statement that I found in my real analysis text book. How can I prove the above? Should I use the theorem:

If $\left\{x_n\right\}$ and $\left\{y_n\right\}$ are two convergent sequences and there exists a natural number $m$ such that $x_n>y_n$ for all $n\geq m$, then $\lim x_n\geq \lim y_n$.

Please anyone help me. Thanks in advance.

Answer 1

You can use the theorem you mention: define $\;\{a_n\}=\{a,a,a,...\}\,,\,\,\{b_n\}=\{b,b,b,...\}\;$ , two constant sequences. Then trivially

$$a_n\le x_n\le b_n\stackrel{\text{by the theorem}}\implies\;\;a=\lim_{n\to\infty}a_n\le\lim_{n\to\infty}x_n=c\le\lim_{n\to\infty}b_n=b$$

Answer 2

Sure, and note that your result is also true in $x_n\geq y_n$. For your problem:

You know that, in fact, for all $n$, $a\leq x_n\leq b,$ and so then $a\leq \lim x_n\leq b.$

Answer 3

You shouldn't have to use any theorems, it follows directly from the definition of convergence of a sequence. Assume that $c$ is not in $[a,b]$. Then either $c<a$ or $c>b$. I will do the proof for $c<a$. Let $\epsilon = \frac{a-c}{2}$. Since $x_n\to c$, $\exists N$ such that $\forall n>N$, $x_n-c<\epsilon$. But $x_n \geq a$ so $x_n-c\geq a-c=2\epsilon>\epsilon$, a contradiction. The other case is very similar, can you do it?

Answer 4

Let's assume that $ c> b $ with $\lim_{n\to\infty}x_n=c$ and arrive at a contradiction. We have $$ (\forall \epsilon>0)(\exists N_\epsilon\in\mathbb{N})(\forall n\in\mathbb{N})\big[(n>N_\epsilon)\implies(c-\epsilon<x_n<c+\epsilon)\big] $$ Choose $ \epsilon = \frac{1}{2}(c-b)>0 $. Then there is $N_{\frac{1}{2}(c-b)}$ such that for all $n>N_{\frac{1}{2}(c-b)}$ we have $$ c-\frac{c-b}{2}<x_n< c+\frac{c-b}{2} \iff b<\frac{c+b}{2}<x_n<\frac{3c+b}{2} $$ But this contradicts the fact that $$ a<x_n<b \quad \forall n\in\mathbb{N} $$ And supposing that $c<a$ we come to an analogous contradiction.