In fact under some set theoretic axioms we can prove that $\Bbb R$ is a set of size $\aleph_2$. So most of those $\aleph_2$ functions are very ragged, and their graphs do not look like anything you would call a "curve". If you think about the cardinality of the real numbers as $\aleph_1$, then you might as well think about the cardinality of $\mathcal P(\Bbb R)$ as $\aleph_2$. Making statements based on opinion; back them up with references or personal experience. You are using standard terminology incorrectly. :). Using this definitions, we have really standard ways to define $\aleph_1$. ...so for example In 2001, I showed that the formulations of OCA of Abraham-Rubin-Shelah and of Todorcevic taken together imply that the continuum is $\aleph_2$. has cardinality $\aleph_2$. The continuum isn't provably a set of cardinality $\aleph_1$ unless you assume the continuum hypothesis. If $f(x)=g(x)$ for all $x\in\mathbb Q,$ then $\{x:h(x)=0\}$ is a closed set containing $\mathbb Q,$ i.e., it's the whole real line, since $\mathbb Q$ is dense in $\mathbb R.$. How do I legally resign in Germany when no one is at the office? Can someone be saved if they willingly live in sin? Consider the set of all well-orderings of $\Bbb N$, namely $X=\{R\subseteq\Bbb{N\times N}\mid (\Bbb N,R)\text{ is a well order}\}$. @Hayden: Sure, I could have put loads more information in about the continuum, but the OP seems confused enough! It might be $\aleph_1$ and $\aleph_2$, but $\mathfrak{c}=\aleph_1$ is the CH. @AsafKaragila Thank you vey much! Can there be an $\aleph_2$ or even greater? @user192369: Not that there is anything wrong with Clive's answer; but there is nothing more annoying than spending time writing an answer to a question, asked merely 20 minutes ago, and upon finishing finding out that the first answer has been accepted. How does the Dissonant Whispers spell interact with advantage from the halfling's Brave trait? @HighGPA If $f:\mathbb R\to\mathbb R$ is continuous, then for any real number $x_0$ there is a sequence of rational numbers $r_n$ such that $x_0=\lim_{n\to\infty}r_n$ and therefore, by continuity, $f(x_0)=\lim_{n\to\infty}f(r_n).$, @HighGPA If $f,g:\mathbb R\to\mathbb R$ are continuous functions, then $h=f-g$ is a continuous function, and so $\{x:h(x)=0\}$ is a closed set. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. He famously showed that the set of real numbers is uncountably infinite. This is all still new territory for me. Can you have a Clarketech artifact that you can replicate but cannot comprehend? The most canonical set of cardinality $\aleph_{\alpha}$ (for any ordinal number $\alpha$) is the ordinal $\omega_{\alpha}$. The second sentence states this correctly. To learn more, see our tips on writing great answers. Oh, of course. $$\omega_1 = \text{the set of countable ordinals}$$ By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Hint: First convince yourself that $|\mathbb R|=|[0,1)|$; every real in $[0,1)$ can be written in base $2$ as an infinite binary string.Second, $2^{\mathbb N}$ is the set of infinite binary strings. if I did? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. You need set theory (or something similar) to do (1), but not (2). Silly of me. You can prove, using set theory, that there are an infinite number of axioms of ZFC. Or perhaps you meant to say $ \mathcal P ( \mathcal P ( \mathbb N ) ) $. The cardinality of the set of real numbers (cardinality of the continuum) is $${\displaystyle 2^{\aleph _{0}}}$$. Asking for help, clarification, or responding to other answers. $$\omega_2 = \text{the set of ordinals whose cardinality is} \le \aleph_1$$ Card$(\bigcup\limits_{n\in\omega} \aleph_2\times\dots\times\aleph_2)=\aleph_2$. Once again, you're confusing (1) proving things about first order logic with (2) using first order logic. What is this hole above the intake of engines of Mil helicopters? The definition of $\aleph_2$ is the least uncountable cardinal larger than $\aleph_1$. In Star Trek TNG Episode 11 "The Big Goodbye", why would the people inside of the holodeck "vanish" if the program aborts? The continuum, i.e. The symbols do not mean what you think. @user192369: Proof that $|\Bbb R|=\aleph_2$ depends on additional axioms. Georg Cantor introduced the concept of cardinality to compare the sizes of infinite sets. Now if we consider an ordinal as a set $T$ such that $(T,\in)$ is a well ordered set, and whenever $x\in T$, then $x\subseteq T$ (for example $\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}$ are all ordinals). :-). 2. Now consider $\equiv$ defined on $X$ as the order-isomorphism relation. I heard a professor refering to a curve with discontinuities in space (R^3) but didn't quit get it.. Any ideas? This talk will discuss some partial results and test questions. Proof that the set of all possible curves is of cardinality $\aleph_2$? :P, That is true, just thought it might be worth mentioning. Canada, Commercial and Industrial Mathematics Program, Centre for Quantitative Analysis and Modelling, Dean's Distinguished Visiting Professorship, CAIMS-Fields Industrial Mathematics Prize, Mathematics-in-Industry Case Studies Journal. Is Firefox so insecure it's worth blocking. Small typo : Line one should read $ \aleph_2 = |\mathcal P(\mathbb R)| $ , not $ | \mathcal P ( \mathcal P (\mathbb R )) $. Though it is provably unequal to some values, like $\aleph_\omega$. Thanks. If you think of "the continuum as an example for cardinality $\aleph_1$" then I'm guessing that you (or your professor) are (at least implicitly) assuming the so-called Generalized Continuum Hypothesis. Thanks for contributing an answer to Mathematics Stack Exchange! Instead, the definition of $\aleph_1$ is the least cardinality larger than $\aleph_0$. Of course, the way I wrote the definitions here require us to assume the axiom of choice, but this can be slightly modified to avoid this issue. However, it is important to realize that we are considering arbitrary functions. But that doesn't mean that you need set theory in order to say what the axioms of ZFC are. @Nick: I watch so many TV shows and movies, that my slang cannot be not fine. Understanding the density operator in quantum mechanics for a joint system, Understanding the mechanics of a satyr's Mirthful Leaps trait. The question of whether Todorcevic's formulation of OCA settles the value of the continuum remains an open problem. Of course, as Clive pointed out, it is provably unprovable that $|\Bbb R|=\aleph_1$ from the standard axioms of set theory (namely $\sf ZFC$), so it is also unprovable that $\mathcal P(\Bbb R)$ has size $\aleph_2$. It cannot be determined from ZFC (Zermelo–Fraenkel set theory with the axiom of choice) where this number fits exactly in the aleph number hierarchy, but it follows from ZFC that the continuum hypothesis, CH, is equivalent to the identity Why would I choose a bike trainer over a stationary bike? @HighGPA Since a continuous function on $\mathbb R$ is determined by its values on $\mathbb Q$ (the set of all rational numbers), the number of continuous functions $f:\mathbb R\to\mathbb R$ is no greater than the number of functions from $\mathbb Q$ to $\mathbb R$ which is $$|\mathbb R^\mathbb Q|=|\mathbb R|^{|\mathbb Q|}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$ which is equal to $\aleph_1$ since we're assuming the continuum hypothesis. Aleph 2, of Cantor's infinite sets X0