Strongly decreasing sequence of sets

This was a homework question few weeks ago, and the teacher didn't give

out the solution to this problem (it doesn't directly concern complex analysis), so I was wondering weather someone knows the answer to this one.

Let $\displaystyle A\subset \mathbb{C}$, and $\displaystyle D$ be such that

$\displaystyle D(A)$ is the set $\displaystyle A$ with all isolated points of $\displaystyle A$ removed. So the question is, is there a set $\displaystyle A\subset \mathbb{C}$

such that the sequence, $\displaystyle (D^{n}(A))_{n\in\mathbb{N}}$ is strongly decreasing?

That is $\displaystyle D(A)\subset A$, $\displaystyle D^{2}(A)\subset D(A)$, but not equal?

My first thoughts went something like this: Let $\displaystyle M:=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}$. Then $\displaystyle D(M)=\{0\}$ and

$\displaystyle D^{k}(M)=\emptyset$ if $\displaystyle k\geq 2$, so this wont do.

Now for every $\displaystyle x\in M$ $\displaystyle x\ne0$ we can make $\displaystyle \varepsilon$-balls such that $\displaystyle \varepsilon_{x_1}\cap\varepsilon_{x_2}=\emptyset$ if $\displaystyle x_1\ne x_2$ and neither is 0. Then we can construct a sequence

for every $\displaystyle x\in M$ such that its limit is $\displaystyle x$ and it is entirely contained in $\displaystyle \varepsilon_x$. Lets call the set of those sequences $\displaystyle S_M$.

Now let $\displaystyle M_1=M\cup S_M$.

$\displaystyle M_1$ is such that $\displaystyle D(M_1)=M$ so now we got 4 strictly decreasing items in the sequence.

For every given $\displaystyle n\in\mathbb{N}$ we can continue this construction

such that the sequence is strictly decreasing for the first $\displaystyle n$ terms.

But that is not the same as the infinite sequence is strictly decreasing.

I am starting to think that this is impossible, but I am not sure.

Does anyone have any ideas?