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We know that $\aleph_0$ is smaller than $\mathfrak c$, the cardinality of the continuum. But are there some good upper-bounds?

For example, it is trivial that $\mathfrak c<2^{\mathfrak c}$, but I wonder if there are better bounds. Specifically, I have to wonder if there exists $\alpha\in\mathbb N$ such that we know that $\mathfrak c<\aleph_\alpha$? If not, what about transfinite ordinals $\alpha$?

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    @chelivery I don't think I want to assume the continuum hypothesis here... – 2017-02-05
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    Sorry, deleted my comment as I read your answer a bit more carefully. Still, generalized CH poses that $\aleph_{n+1}$ is the "next" infinite cardinal after $\aleph_n$, and so $\aleph_2$ would be the bound you're looking for – 2017-02-05
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    @chelivery GCH is even worse. I'm honestly asking you not to have CH, or else there is no point in this question. – 2017-02-05
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    A clarification of my comment, which isn't entirely correct: generalized CH states that $2^c = \aleph_2$. The fact that $\aleph_{n+1}$ is the first infinite cardinal bigger than $\aleph_n$ is just the property of the aleph numbers. So the bound would be $2^c$. – 2017-02-05
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    Also note that if you don't want to involve the continuum hypothesis you have the same open question about the lower bound. – 2017-02-05
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    There is no upper bound in ZFC. See https://en.wikipedia.org/wiki/Cardinality_of_the_continuum "[The cardinality of the continuum can be] most other alephs, although in some cases equality can be ruled out by König's theorem on the grounds of cofinality, e.g., $\mathfrak{c} \neq\aleph_\omega$. In particular, $\mathfrak{c}$ could be either $\aleph_1$ or $\aleph_{\omega_1}$, where $\omega_1$ is the first uncountable ordinal, so it could be either a successor cardinal or a limit cardinal, and either a regular cardinal or a singular cardinal." – 2017-02-05
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    @chelivery I don't believe you understand what I'm saying. It is not necessarily true without CH. The lower bound is indeed independent of CH...see above – 2017-02-05
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    @ziggurism Ah, ok. Thanks for that. – 2017-02-05
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    $\aleph_0 < c$ is independent of CH but you are asking for a "better upper bound". That $\aleph_0$ is the best lower bound is precisely the continuum hypothesis, that's what I mean with the "same open question". Sorry if I misinterpreted your question, though! – 2017-02-05
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    @chelivery The OP is asking for a better upper bound on the size of $\mathfrak c$ than $\mathfrak c < 2^{\mathfrak c}$, for instance some $\kappa$ such that $\mathfrak c < \aleph_\kappa$. (There isn't one.) – 2017-02-05
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    @SimplyBeautifulArt So I think it is consistent with ZFC for $\mathfrak{c}$ to be any cardinal $\kappa$ with $\operatorname{cf}(\kappa)>\aleph_0$ – 2017-02-05
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    @ziggurism Ok, thank you for your comments! :D – 2017-02-05
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    @SimplyBeautifulArt: Sure. I also noticed Andrés Caicedo's answer here: http://math.stackexchange.com/a/20387/16490 – 2017-02-05

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