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

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

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