The definition Im reading:
Well, say I had the open set A. Then by definition, if a is an element of A, then the open interval (a- $\epsilon, a+\epsilon$) is also a part of A. So if a+$\epsilon$ is a part of a, then since A is the open set, a+2$\epsilon$ must be a part of A as well. So on and so forth, we get that every number is a part of A. So If A is an open set, then A is a subset of the Reals and the Reals is a subset of A?
What's wrong with this interpretation?
On
More than that: The 2 above comments used the notation there exists some $\varepsilon$ for which $ (a-\varepsilon,a+\varepsilon)\subset A$. I would like to add to that:
a. The point $a+\varepsilon$ need not be in $A$. b. for every $0\leq r<\varepsilon$ $[a-r,a+r]\subset A$
In my opinion, it is not only that claim is true for some $\varepsilon$ it is also that the boundary of the interval $ (a-\varepsilon,a+\varepsilon)$ need not be in A.
On
For every $a\in A$ there exists an $\epsilon>0$ such that...
Element $a$ has "some authority" over this $\epsilon$. It determines how large this $\epsilon$ can be.
You could pick one out and denote it as $\epsilon_a$.
Then $(a-\epsilon_a,a+\epsilon_a)\subseteq A$ and taking $\epsilon_a$ small enough we can even have $[a-\epsilon_a,a+\epsilon_a]\subseteq A$.
Then indeed $b:=a+\epsilon_a\in A$, and you can also find an $\epsilon_b>0$ with $a+\epsilon_a+\epsilon_b=b+\epsilon_b\in A$.
But possibly you are forced to take $\epsilon_b<\epsilon_a$ so you cannot conclude that $a+2\epsilon_a\in A$.
edit:
You are missing that it says "there exists" an $\varepsilon$. So you cannot just pick any $\varepsilon$ regardless of what $a$ you pick.