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.
 A: 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, …").
A: 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.
A: 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".
