# Special chains in $\mathbb{R}^n$

We had a topology exam yesterday, where the following was a question:

Prove or disprove that there is a chain (ordered by set inclusion) of discrete subsets of $\mathbb{R}^n$ containing uncountably many elements. By a discrete set, one means a set in which every point is an isolated point. To clarify, each element of the chain should be a discrete subset of $\mathbb{R}^n$, and the chain itself should contain countably/uncountably many elements.

I am very curious as to what the answer is. Intuitively, it seems to me that the largest such chain should consist of countable elements, but I have no idea how to proceed with the proof. I will be grateful for any ideas.

• The question is ambiguous. It is not clear what is supposed to contain uncountably many elements. It is the chain (so the elements are the discrete subsets), or is it each of the discrete subsets (so the elements are point of $\mathbb R^n$)? In other words, should we read "(chain of sets) with uncountably many elements" or "chain of (sets with uncountably many elements)"? – MPW Jun 30 '16 at 14:48
• @MPW Sorry about the ambiguity, have edited the question. – user351219 Jun 30 '16 at 14:57

Note that $\mathbb{N}^n$ is an infinite discrete subset of $\mathbb{R}^n$, and therefore, so too is any subset. Hence, if we can find an uncountable chain in $\mathcal{P}(\mathbb{N}^n)$ we are done.
As $\mathbb{N}^n$ is countable and we no longer care about any interal structure, we may as well replace it with any countable set we like, and look for a chain in its powerset. In particular, we may as well look for a chain in $\mathcal{P}(\mathbb{Q})$. But then, each $r\in \mathbb{R}$ determines a unique subset of $\mathbb{Q}$: $$D_r = \{ q\in\mathbb{Q} : q<r\}$$ Clearly if $r_1 < r_2$ then $D_{r_1}\subsetneq D_{r_2}$. So then $\{D_r : r\in \mathbb{R}\}$ is a chain in $\mathbb{Q}$ of size continuum, and so we are done.
The above construction is the standard Dedekind cut method of constructing $\mathbb{R}$ from $\mathbb{Q}$.
• I am a bit confused, are the sets $D_r$ discrete? No point in $D_r$ seems to be isolated. – user351219 Jun 30 '16 at 14:44
• @user351219 This is where we are using a trick (the part where I said "we may as well replace it with any countable set we like"). Formally, pick a bijection between $\mathbb{N}^n$ and $\mathbb{Q}$. We've built a chain in $\mathcal{P}(\mathbb{Q})$, and, using $f$ you can pull that back to a chain in $\mathcal{P}(\mathbb{N}^n)$. Every subset of $\mathbb{N}^n$ is discrete, and so we are done. – James Jun 30 '16 at 14:47