Confuse about definition of dense set

Question

Assume (M,d) is a metric space , A is a subset of M

"A is dense in M " means $\forall$ x$\in$ M ,we always can find points of A to approach x

My question is ：

Since we are using the concept of "approach" to define the dense set,why not use A'= M as the definition of "A is dense in M", instead use cl(A) = M as its definition?

PS: A'= the derived set of A and cl(A) is the closure of A

Answer 1

The difference is in what happens at isolated points.

Consider for example a discrete metric space like $M=\mathbb{Z}$. If we take $A$ to be $\mathbb{Z}$ itself, should we not say that $\mathbb{Z}$ is dense in $\mathbb{Z}$? After all, it is possible to "approach" all the points in $\mathbb{Z}$ (extremely well!) by points of $\mathbb{Z}$, namely themselves. (You can "approach" 0 by the sequence $(0,0,0,0,0,\dots)$.

But according to your proposed definition, $\mathbb{Z}$ would not be dense in $\mathbb{Z}$, since the derived set of $\mathbb{Z}$ is empty.

So your definition is not equivalent to the usual definition of "dense". One could argue about which definition better deserves the name of "dense", but the majority has already spoken and is not going to change their minds now.

Answer 2

If you define $A$ is dense iff for all $x \in M$ there exists a sequence $(a_n)$ from $A$ such that $a_n \to x$ then this is the same (in metric spaces) as demanding $\operatorname{cl}(A)=M$.

$A' = M$ need not hold, as $M$ can have isolated points that are not in any derived set. But if $\{x\}$ is an open set and $A$ is dense, then $x$ must lie in $A$ and we can take a constant sequence $x$ to "approximate" $x$ by a sequence. We can only have a set $A$ with $A' =M$ when $M$ has no isolated points, so this is not taken as a general definition of being dense for all metric spaces.