Let $\{x_n\}$ be a sequence such that $a\leq x_n\leq b$. Suppose the sequence converges to $x$. Show that $a\leq x\leq b$.

My attempt:

Take $\epsilon>0$.

There is an $N_0$ such that $a\leq x_n\leq b$ when $n\geq N_1$.

There is an $N_1$ such that $a< \epsilon$ when $n\geq N_1$.

There is an $N_2$ such that $b<\epsilon$ when $n\geq N_2$.

Selecting $N=\max(N_0,N_1,N_2)$. Then when $n\geq N$, $-\epsilon<a\leq x_n-x\leq b < \epsilon$, and $a\leq x\leq b$.

Is this the right approach?

  • $\begingroup$ No. You cannot use $a<\epsilon$, $b<\epsilon$. Youmight have $a=1$, $b=2$, for example, but $\epsilon$ is small $\endgroup$ – Hagen von Eitzen Sep 29 '16 at 6:00
  • $\begingroup$ Please. Look at my edits to the question. Where in the world did you get your ideas of how to edit MathJax code? $\qquad$ $\endgroup$ – Michael Hardy Sep 29 '16 at 6:17
  • $\begingroup$ @MichaelHardy. I appreciate your edit. Today is my first time using it. $\endgroup$ – ozarka Sep 29 '16 at 6:19

By definition of limit for all $\epsilon>0$ there is $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, $$x-\epsilon<x_n< x+\epsilon.$$ Now assume that $x>b$ and let $\epsilon=(x-b)/2>0$, then $$b<\frac{x+b}{2}=x-\frac{x-b}{2}<x_n\leq b$$ which is a contradiction.

What happens when $x<a$?

  • $\begingroup$ Unfortunately I can't see it. I'm assuming that I need to show the inequality will not hold when $x>b$? All I've done so far is substitute $\epsilon$ with your formulation. I've just been stuck, staring at it. $\endgroup$ – ozarka Sep 29 '16 at 6:22
  • $\begingroup$ I can say that: $x$ - $(x-b)$/$2$ $<$ $x$ $+$ $(x-b)$/2 to arrive at $-2x$<$-2b$ which implies $x$ $<$ $b$ but we assumed the opposite. So $b$ must be greater than $x$. Now should I assume by contradiction that $x$<$a$ and $\epsilon$ = $-$$(x-a)$/$2$ $>$ $0$? $\endgroup$ – ozarka Sep 29 '16 at 6:29
  • 1
    $\begingroup$ @ozarka No. You have that $b<x-(x-b)/2<x_n\leq b$. Try the case $a>x$. $\endgroup$ – Robert Z Sep 29 '16 at 6:51
  • $\begingroup$ so I assumed that $x$<$a$. and let $\epsilon$ = $-$$(x-a)$/$2$>$0$. Then $a$ > $(x+a)/2$ = $x$ + $-$$(x-a)$/$2$ < $x_n$ $\geq$ $a$. A contradiction. ? $\endgroup$ – ozarka Sep 29 '16 at 6:55
  • 1
    $\begingroup$ @ozarka Yes, it is correct. $\endgroup$ – Robert Z Sep 29 '16 at 7:01

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