What is the difference between a dense set and an $\varepsilon$-dense set?

Question

I'm reviewing for an upcoming exam for my metric spaces class and I am confused on two definitions I was provided. Here are said definitions as per my notes:

1. Recall: Let $(X, d)$ be a metric space. $A \subseteq X$ is dense means that $\overline{\rm A} = X$ or in other words is $\forall \varepsilon > 0$, $\forall x \in X$ $\exists y \in A$ such that $d(x,y) < \varepsilon$.

2. $A \subseteq X$ is called $\varepsilon$-dense if $\forall x \in X$, $\exists y \in A$ such that $d(x,y) \leq \varepsilon$.

Is the difference between these two concepts just one being something about two points being stricly less than $\varepsilon$ and one being $\leq \varepsilon$? Why would we care about something being $\varepsilon$-dense versus just normal dense? Isn't something being normal dense a stronger condition? What is the motivation for using the latter definition over the former?

Thanks! :)

Answer 1

A set $A$ is dense if every point of $X$ is a limit point of $A$; meaning that for any $x \in X$, you can get arbitrarily close to $x$ with points in $A$.

If $A$ is $\epsilon$-dense, it means you cannot get arbitrarily close, but rather only that given $x \in X$, you can get within $\epsilon$ (for a fixed $\epsilon$) of this point $x$ from some point $a \in A$.

The concept of $\epsilon$-dense can be thought of as a kind of resolution problem; if you are $\epsilon$-dense, you are "dense up to scale/resolution $\epsilon$".

Answer 2

Exercise. Prove in a metric space, A is dense iff
for all r > 0, A is r-dense.

Dense is a topological property,
e-dense is only a metric property.