Is $2^{\aleph_0} = \aleph_1$?

I was reading a thread on Examples of Common False Beliefs in Mathematics on MathOverflow, in which a user wrote:

$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

But I thought that $\aleph_0 = |\Bbb{Z}|$ and $\aleph_1 = 2^{\aleph_0} = |\Bbb{R}|$. I'm surprised to hear the opposite asserted. Who is right, and for what reasons?

Note: I haven't had sufficient mathematical exposure to prove anything about cardinals, but I'd like to understand what's going on here.

• you can read about the continuum hypothesis : en.wikipedia.org/wiki/Continuum_hypothesis – Tryss Apr 17 '15 at 7:04
• $\aleph_1$ is defined to be the cardinality of the set of countable ordinal numbers. We have $2^{\aleph_0} = \aleph_1$ iff the continuum hypothesis is true. – William Stagner Apr 17 '15 at 7:05
• Thanks for the references -- I did not know about the continuum hypothesis. – Newb Apr 17 '15 at 7:16
• – Martin Sleziak Apr 17 '15 at 7:18
• It is truly a common false belief. (Unlike some of the others in that thread.) It has appeared in "popular" math books by non-mathematicians. – GEdgar Apr 17 '15 at 21:42

The cardinality of real numbers is indeed $2^{\aleph_0}$. However $\aleph_1$ is not defined as $2^{\aleph_0}$ (the cardinality of the continuum, often denoted $\mathfrak c$), but it is defined as the smallest cardinality strictly greater than $\aleph_0$. Whether $2^{\aleph_0}=\aleph_1$ (termed the continuum hypothesis) can neither been proven nor disproven from ZFC. You can of course add it as additional axiom to ZFC, and in that extended set theory, it is then indeed true. However you can also add the axiom $2^{\aleph_0}\ne\aleph_1$ (that is, the assumption that there exists a set whose cardinality lies strictly in between the natural and the real numbers) to ZFC and get another set theory in which the claim is false.

Now usually mathematicians assume ZFC, not ZFC+continuum hypothesis (nor ZFC+negation of the continuum hypothesis), therefore it is a false belief that this relation must be true.

It is however not a false belief that this relation is true; that is only an unprovable belief. There's nothing inconsistent with assuming it to be true; you cannot disprove it; however you also cannot prove it. But that's true of many believes.

However you cannot use it in a proof unless you explicitly specify it as precondition of what you want to prove ("assuming the continuum hypothesis is true, …").

• If we took the continuum hypothesis as an additional axiom , what additional statement that were independent of zfc can be proven? And what if we took negation of continuum hypothesis as an axiom? – A Googler Apr 17 '15 at 7:35
• math.stackexchange.com/q/178069/34930 – celtschk Apr 17 '15 at 7:49
• Intuitively, can I imagine this is similar to the fifth postulate of Euclidean geometry? – MonkeyKing Apr 20 '15 at 2:18
• @MonkeyKing: Indeed, there's a parallel (no pun intended): It was tried to prove it from the other axioms for some time before it was shown to be independent. However I don't think it was ever considered to be self-evident as the fifth postulate was for millennia. – celtschk Apr 20 '15 at 18:38

Essentially, people are confusing $\aleph_1$ with $\beth_1$ (pronounced "beth", see beth numbers).

More precisely: People have heard of the $\aleph$ numbers and cardinal exponentiation and think $\aleph_1=2^{\aleph_0}$ which is not true (unless we assume an extra axiom which "by default" we do not). However, people have not heard of the $\beth$ numbers, and their definition:

$$\beth_{x+1}=2^{\beth_x}$$

seems to be just what some people think the $\aleph$ numbers are.

There are good reasons that $2^{\aleph_0}=\aleph_1$ (the continuum hypothesis) is probably true, even if ZFC (+ large cardinals) alone cannot prove it. So people may not be too wrong when naively believing in this proposition.

However, we cannot yet be sure, this is still ongoing research. You may want to look at Hugh Woodin's work on "Ultimate-L".

• Probably true where? – Asaf Karagila Apr 17 '15 at 20:56
• Out of curiosity, what are these "good reasons" you talk about? From what I've heard (I can't reference where I've heard that), the majority would rather believe CH to be false, not true. – Wojowu Apr 17 '15 at 21:27
• True from a platonistic point of view, in the sense of "really true", just like the axioms of ZFC and large cardinal axioms (and projective determinacy and so on) are true facts about the "real" mathematical universe. – Gerald Apr 17 '15 at 21:30
• @Wojowu: You are right that until about the end of the 20th century most logicians felt that CH should be false (Cohen: "The power set operation is quite powerful"). But this has changed. As mentioned, search for "Woodin" and "V=Ultimate-L". – Gerald Apr 17 '15 at 21:34