Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $x \in H_{\aleph_1}$, then $L_{\omega_1}(x) \subset H_{\aleph_1}$.

Here, what exactly is $L_{\omega_1} (x)$? And can anyone show an example of how to form hereditarily countable set?

share|cite|improve this question
up vote 1 down vote accepted

In order to make sense of $L_{\omega_1}(x)$, you should first have a basic understanding of $L$, the constructible universe; the description here is as understandable a short description as you’re likely to find. The same idea is used to define $L_\alpha(x)$ for a set $x$, the universe of sets constructible from $x$; it’s described here.

A set $x$ is hereditarily countable if it is countable, and each of its elements is hereditarily countable. Alternatively, $x$ is hereditarily countable if it is countable, and so is each element of its transitive closure. The countable ordinals are handy examples of hereditarily countable sets.

The idea behind the theorem that you quote is that if you start with a hereditarily countable set, you can define only countably many things from it in one step, so each $L_\alpha(x)$ for $\alpha<\omega_1$ will still be countable, and indeed hereditarily so.

share|cite|improve this answer

The constructible universe is denoted by $L$, and it is constructed by a transfinite induction, where the $\alpha$-th stage is denoted by $L_\alpha$.

We can talk about relative constructibility, which is the same construction only we start with a transitive set which may not be the empty set as in $L$. We use $L(x)$ to denote the sets constructed in this fashion when we began with the transitive closure of $x$.

To say that $x\in H_{\omega_1}$ is to say that $x$ has a countable transitive closure. And $L_{\omega_1}(x)$ is the set of all those which were constructed at countable levels from the transitive closure of $x$.

To show this, prove by induction that if $x\in H_{\omega_1}$ then $L_\alpha(x)$ is countable for every countable ordinal. Since $L_\alpha(x)$ is always a transitive set, and it is countable, it must be a subset of $H_{\omega_1}$, and so must be $L_{\omega_1}(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.