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 $[a, b]$.

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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 &lt; a, then there exists some \epsilon &gt; 0 such that x \notin (a - \epsilon, b + \epsilon). For example, if \epsilon = (a - x)/2 then a - \epsilon = (a + x)/2 &gt; x.

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

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