Why <HF,⊆> is simply isomorphic to the lattice of finite subsets of a countable set, such as the collection of finite subsets of N under inclusion.

Question

< <HF,⊆> is simply isomorphic to the lattice of finite subsets of a countable set, such as the collection of finite subsets of N under inclusion. why?Can we have <HF,⊆> is simply isomorphic to finite subsets of N under devide relation <= to <N,|>

Answer 1

Suppose $X$ is a countable set, and let $[X]^{<\omega}$ be the set of finite subsets of $X$. Consider a bijection $f: X\rightarrow HF$. This bijection induces a map $$\hat{f}: [X]^{<\omega}\rightarrow [HF]^{<\omega}$$ given by $$A\mapsto \{f(a): a\in A\}.$$ Moreover, $\hat{f}$ "doesn't change subsethood" in the sense that $$A\subseteq B\iff \hat{f}(A)\subseteq \hat{f}(B)$$ for all $A,B\in[X]^{<\omega}$. Put another way, $\hat{f}$ is an isomorphism between the structures $([X]^{<\omega},\subseteq)$ and $([HF]^{<\omega},\subseteq)$.

• Note that $\hat{f}$ is just a restriction to finite subsets of the usual function $\mathcal{P}(X)\rightarrow\mathcal{P}(HF)$ induced by $f$.

But here's the cute bit: it turns out that $HF=[HF]^{<\omega}$. This may look weird at first and it's certainly not true that $X=[X]^{<\omega}$ in general - it's a special feature of $HF$. So we have in fact $$([X]^{<\omega},\subseteq)\cong([HF]^{<\omega},\subseteq)=(HF,\subseteq).$$ (That "$=$" instead of "$\cong$" isn't a typo, they are literally the same object.)