Difference between $(0, 1)$ and $[0 + \epsilon, 1 - \epsilon]$ I'm wondering if there's any real difference between $(0, 1)$ and $[0 + \epsilon, 1 - \epsilon]$. Aren't they just the same? I mean, don't they contain the same numbers? I know the second is closed and the first is not, but I am talking in terms of numbers.
 A: An important thing to keep in mind is that $\epsilon$ is a number just like any other. It just so happens that it's frequently used in a context where you are considering a sequence of different $\epsilon$ tending toward $0$. Let me give a couple examples.
If you write only $[0 + \epsilon, 1 - \epsilon]$ for fixed $\epsilon > 0$, then this is a proper subset of $(0,1)$ because $\epsilon$ is a fixed constant. For example, if $\epsilon = 0.1$, then we're just talking about the interval $[0.1, 0.9]$.
What you presumably have in mind is something like
$$
\lim_{\epsilon \rightarrow 0^+} [0 + \epsilon, 1 - \epsilon]
$$
or
$$
\bigcup_{0 < \epsilon < 0.5} [0 + \epsilon, 1 - \epsilon],
$$
which is equal to $(0,1)$ because $\epsilon$ is eventually smaller than any real number in $(0, 1)$. In either case, however, we are referring to the limit of a sequence of intervals using different $\epsilon$, not a single fixed $\epsilon$.
A: They are not the same.$$\frac{1}{2}\epsilon\notin[0 + \epsilon, 1 - \epsilon]$$But$$\frac{1}{2}\epsilon\in(0 , 1)$$
