Classification of subspace topologies of countably infinite subsets of $\mathbb{R}$

Question

For example, we know that $\mathbb{Z}$ is discrete, while $\mathbb{Q}$ is not. The former has no limit point, while for the latter, every point is a limit point. An intermediate object is $X_l=\{l\}\cup\{l+1/n\}_{n\in\mathbb{N}}$, which has only one limit point. I conjecture that for the desired classification, it suffices to see which point is a limit point, and that there are $2^{\mathbb{Z}}$ different subspace topologies. The operations such as $X_l\cup\mathbb{Z}$ increase a the number of limit points by one, and they may explicitly construct an arbitrary subspace topology from $\mathbb{Z}$.

Questions

• Are my conclusion and argument correct? If not, please answer the correct conclusion regarding the classification and justify it.
• For this classification, is there an explicit way to construct any such topology from the known objects such as $\mathbb{Z}$ and $\mathbb{Q}$?

Edit Qiaochu showed that limit points are not sufficient for classification. What tools would be needed?

Answer 1

The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $\mathbb{Q}$ as a dense in itself kernel.

The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(\alpha+1)} = (X^{(\alpha)})'$, for successor ordinals $\alpha +1$, and $X^{(\beta)} =\bigcap \{X^{(\alpha)} \alpha < \beta\}$ for limit ordinals $\beta$. This process stops on some countable ordinal $\gamma$ such that $X^{(\gamma)} = X^{(\gamma+1)}$. Then $X$ is uniquely determined by this $\gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $\mathbb{Q}$ essentially.)

So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.

Answer 2

It does not suffice to record which points are limit points. For example, the subspace given by

$$\{ -n, n \in \mathbb{N} \} \cup \{ 0 \} \cup \left\{ \frac{1}{n}, n \in \mathbb{N} \right\}$$

and the subspace given by points of the form

$$\left\{ -\frac{1}{n}, n \in \mathbb{N} \right\} \cup \{ 0 \} \cup \left\{ \frac{1}{n}, n \in \mathbb{N} \right\}$$

both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.