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What is the first cardinal number which is grearter than continuum? We denote it by ? Thanks very much.

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3 Answers 3

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$\mathfrak{c}^+$ or $( 2^{\aleph_0} )^+$; the $^+$ denotes that we are taking the cardinal successor. Its value in the (transfinite) sequence of aleph numbers cannot be determined by ZFC because the value of $\mathfrak{c} = 2^{\aleph_0}$ cannot be determined by ZFC.

The Continuum Hypothesis is the conjecture that $\mathfrak{c} = \aleph_1 = \aleph_0^+$, and was proven to be independent of ZFC by Kurt Gödel (ZFC cannot prove $\mathfrak{c} \neq \aleph_1$) and Paul Cohen (ZFC cannot prove $\mathfrak{c} = \aleph_1$). Furthermore Paul Cohen's proof showed that given any aleph number $\aleph_\alpha$ ZFC cannot prove $\mathfrak{c} \leq \aleph_\alpha$.

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"Cannot be determined" is a little fuzzy here. We can certainly define a concrete set with this cardinality: it's the set of all order-isomorphism classes of well-orderings of the reals. –  Henning Makholm Feb 27 '13 at 10:31
    
@Henning: I was actually just going to expand on this point. –  Arthur Fischer Feb 27 '13 at 10:31
    
Maybe I'm missing something obvious, but it seems to me that by Hartog's construction, there is a cardinal $\kappa$ for which ZFC proves $\mathfrak{c} \leq \kappa$ ? –  us2012 Feb 27 '13 at 13:41
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@us2012: Nope. If I am thinking about the correct Hartog result, he showed that Hartog showed in ZF(C) that given any set $S$ the set of all ordinals injectible into $S$ is a set, and assuming Choice it follows that if $\kappa$ is the least ordinal (cardinal) not injectible into $S$, then $| S | \leq \kappa$. This is fine, however there are no a priori bounds on what $\kappa$ should be (in the $\aleph$-sequence) given $S$; in different models of ZFC you might get different $\kappa$'s. [cont...] –  Arthur Fischer Feb 27 '13 at 14:04
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[..inued] Cohen's method of forcing allows us to start with a model $M$ of ZFC and a $\kappa \in M$ such that $M \models \text{"}\kappa\text{ is an uncountable cardinal"}$ and produce a new model $M^\prime \supseteq M$ in which (1) $M^\prime \models \text{"}\kappa\text{ is a cardinal"}$ for all $\kappa \in M$ such that $M \models \text{"}\kappa\text{ is a cardinal"}$, and (2) $M^\prime \models \text{"}| \mathbb{R} | = \kappa^+ \text{"}$. This means that ZFC could not prove $| \mathbb{R} | \leq \kappa$. –  Arthur Fischer Feb 27 '13 at 14:04

The notation $\beth_1$ represents the cardinality of the continuum. See also, beth numbers.

Thus, by definition, $(\beth_1)^+$ is the first cardinal number which is strictly greater than $\beth_1$. See also, successor cardinal.

Whether $(\beth_1)^+ = \beth_2$ is independent of ZFC. See also, generalized continuum hypothesis.

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The axioms of ZFC can prove that there exists $\alpha$ such that $\frak c=\aleph_\alpha$, but the axioms themselves are insufficient in order to prove much on that $\alpha$. We can prove that $\alpha\neq 0$ and that $\alpha$ does not have cofinality of $\omega$, whatever that might be.

But besides these two facts we can't say anything intelligible on what exactly is $\alpha$. We know it is possible to have a universe of set theory where $\alpha=1$ and another where $\alpha=2$.

Therefore we cannot really say much on what is $\frak c^+$, or $\aleph_{\alpha+1}$, since we don't know what is $\aleph_\alpha$ in this case. We can, however, prove that it exists and give it a symbol such as:

  • $\left(2^{\aleph_0}\right)^+$;
  • $\frak c^+$;
  • $\left(\beth_1\right)^+$

And so on.

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