# Query about Approximation Property of Suprema

The statement for approximation property of Suprema states that: If E has a finite supremum and $\epsilon > 0$ is any positive number, then there is a point $a \in E$ such that $sup E - \epsilon < a \leq sup E$.

Now, I can rewrite the inequality as follows: $sup E - \epsilon < a \leq sup E< sup E + \epsilon$

Which implies: $sup E - \epsilon < a < sup E + \epsilon$

Implying: $- \epsilon < a - sup E < + \epsilon$

Implying: $\left|a - sup E\right| < \epsilon$

and Implying: $\left|a - sup E\right| = 0$.

This means that $a = sup E$ which is not true since there could be instances where sup E can belong outside the set E.

Could you please tell me what's wrong with my logic? Thanks!

• The point $a$ depends on $\varepsilon$, so you cannot deduce that $|a-sup E|<\varepsilon$ implies $|a-\sup E|=0$. It may be the case that as $\varepsilon \to 0$, that the sequence of points $a$ "leave" the set $E$.
– Matt
Commented Oct 30, 2016 at 16:14
• Oh! I didn't know that $a$ has to depend on $\epsilon$. Commented Oct 30, 2016 at 16:19
• Yes, unless the maximum is attained then it must be the case that $a$ depends on $\varepsilon$. You can draw some pictures to build an intuition here for the set $[0,1)$. If you give me an $\varepsilon$ then I can give you an $a$ which is $\varepsilon$-close to 1. But then you can give me a new $\varepsilon$ which is even smaller so that my $a$ doesn't work anymore.
– Matt
Commented Oct 30, 2016 at 16:23
• That makes sense. Could you combine your comments as your answer? I'll accept that as an answer. I think you are correct. Commented Oct 30, 2016 at 16:26
• Just did. Thanks! Good luck with your studies.
– Matt
Commented Oct 30, 2016 at 16:34

Note that the point $a$ may depend on $\varepsilon$, so you cannot deduce that $|a-sup E|<\varepsilon$ implies $|a-\sup E|=0$.
For example, it may be the case that as $\varepsilon \to 0$, we define a sequence of points $a=a_\varepsilon$ so that $a_\varepsilon$ "leaves" the set $E$. For example, consider the set $[0,1)$. Given the sequence $\varepsilon(n)=1/n$, we can take $a(n) = 1-1/2n$, each of which is $\varepsilon$-close to 1. But, for any given $a(n)$, there will eventually exist a $\varepsilon(N)$ for $N$ large enough so that $a(n)$ will no longer be $\varepsilon$-close.