I have three questions about elementary set teory and i don't figure out how to solve them:

1)Let $X$ a subset of the cardinal number $2^{\aleph_0}$ (seen as an initial ordinal). Is true or false that $X$ or $2^{\aleph_0} \setminus X$ (the complement set of $X$ in $2^{\aleph_0}$) has the order type of $2^{\aleph_0}$ ?

2) Exist a set $X$ such that $X \subseteq X \times X $ and a set $Y$ such that $Y \times Y \subseteq Y$ ??

3)Let b an ordinal number that $\omega^b = b$ (ordinal exponentation). We can conclude that for every $s,t < b$ we have $s+t<b$ ?

Thanks in advance

  • $\begingroup$ Please limit yourself to one (or two which are closely related) questions per thread. $\endgroup$ – Asaf Karagila Jan 17 '14 at 14:45
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    $\begingroup$ (The first question also makes no sense at all.) $\endgroup$ – Asaf Karagila Jan 17 '14 at 14:45
  • $\begingroup$ The reason the first question doesn't make sense is that while cardinals are ordinals, we don't associate them with structure. $2^{\aleph_0}$ is a "size" not "order type", and therefore $2^{\aleph_0}\setminus X$ makes no sense, and it makes even less sense to ask about the order type of that. $\endgroup$ – Asaf Karagila Jan 17 '14 at 14:48
  • $\begingroup$ the cardinal $2^{\aleph_0}$ is seen as an initial ordinal. The set $2^{\aleph_0} \setminus X$ is the complement set of X in $2^{\aleph_0}$. $\endgroup$ – camascers Jan 17 '14 at 14:50

I will only answer question 1. The answer is yes. Either $X$ or its complement must have cardinality $2^{\aleph_0}$. For example, say it's $X$. Then the order type of $X$ is at least $2^{\aleph_0}$, because $2^{\aleph_0}$ is the smallest order type with that cardinality. But the order type cannot exceed $2^{\aleph_0}$, because then $X$ would have a proper initial segment of cardinality $2^{\aleph_0}$, which is absurd since this would be contained in a proper initial segment of $2^{\aleph_0}$.

Edit: Here is an answer to part 2, assuming the sets are required to be non-empty.

A set $Y$ satisfying the requirement can be chosen, for example $Y = V_\omega = \cup_{n \in \omega} V_n$, where $V_0 = \emptyset$ and $V_{n+1} = P(V_n)$.

No non-empty $X$ of the required kind exists, however. Apply the axiom of foundation to the set $X \cup (\cup X)$.


The first answer is yes. Since $2^{\aleph_0}$ is an infinite cardinal, a partition into two sets must include one which has cardinality $2^{\aleph_0}$. Since we consider this as an initial ordinal, every subset which as the same cardinality has the same order type. It is interesting to note the use of the axiom of choice here, for having $2^{\aleph_0}$ as an ordinal to begin with. Without the axiom of choice, it is consistent that $\Bbb R$ can be split into two sets, both of which have cardinality smaller than $\Bbb R$ itself.

The second answer is easily yes. $X=Y=\varnothing$. But it is possible to find infinite sets which satisfy the property $A\subseteq A\times A$.

The third question is also yes, but not as easily as before. If $b=\omega^b$, and $t<b$ then we have that $t+b=b$. Now if $t,s<b$ then $t+s<t+b=b$.

  • $\begingroup$ in the first question if X is the empty set we have that $2^{\aleph_0} \ \emptyset = 2^{\aleph_0}$ has the same order type of $2^{\aleph_0}$ if $X^c$ is empty it implies that $X$ is $2^{\aleph_0}$ that has the same order type of $2^{\aleph_0}$ (i'm wrong?) while if the sets in the second questions are not empty, do you know some examples of sets that satisfy that proprieties? $\endgroup$ – camascers Jan 17 '14 at 15:10
  • $\begingroup$ Ohhh, sorry I read that only the complement has order type such and such. $\endgroup$ – Asaf Karagila Jan 17 '14 at 15:19
  • $\begingroup$ To make the question more interesting, why don't you add a small comment on whether choice matters in question 1 (that is, if $\mathbb R$ is split into two sets, must one have size $\mathfrak c$?). $\endgroup$ – Andrés E. Caicedo Jan 17 '14 at 15:37
  • $\begingroup$ @Andres: Well, choice matters for making $2^{\aleph_0}$ an ordinal to begin with... :-) $\endgroup$ – Asaf Karagila Jan 17 '14 at 15:38
  • $\begingroup$ Which is why I only mention size. $\endgroup$ – Andrés E. Caicedo Jan 17 '14 at 15:39

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