Analysis question regarding the relationship between open and closed intervals

[Arkuski]

Let a and b be real numbers with a<b, and let x be a real number. Suppose that for each ε>0, the number x belongs to the open interval (a-ε, b+ε). Prove that x belongs to the interval [a, b].

 

[SammyS]

Let $a$ and $b$ be real numbers with $a<b$, and let $x$ be a real number. Suppose that for each $\epsilon >0$, the number $x$ belongs to the open interval $(a-\epsilon , b+\epsilon )$. Prove that $x$ belongs to the interval $[a, b]$.

Use double $ signs for LaTeX . Double # signs for inline LaTeX .

What have you tried?

Where are you stuck?

 

[Arkuski]

I tried showing that if y was in the compliment of the set then the closed interval would follow. For example, y≤a-ε and y+ε≤a. Since ε>0, we know that y<a. Thus x≥a. I used a similar proof for the upper bound. Is there a better way to do this?

 

[pasmith]

I think your argument works, but the one which certainly works is this:

If $x < a$, then there exists some $\epsilon > 0$ such that $x \notin (a - \epsilon, b + \epsilon)$. For example, if $\epsilon = (a - x)/2$ then $a - \epsilon = (a + x)/2 > x$.

Similarly, if $x > b$ then $x \notin (a -\epsilon, b + \epsilon)$ when $\epsilon = (x - b)/2 > 0$.

Hence if $x \in (a -\epsilon, b + \epsilon)$ for all $\epsilon > 0$ then $x \geq a$ and $x \leq b$. Thus $x \in [a,b]$.