continuum is aleph 2

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 $\mathbb R.$. Understanding how an index is converted to a logarithm. 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 $\aleph_2$. Namely, the least size of an uncountable set. Why right handed circular polarization gets lagged when going through ionized plasma? Or perhaps you meant to say $ \mathcal P ( \mathcal P ( \mathbb N ) ) $. Understanding the density operator in quantum mechanics for a joint system, Understanding the mechanics of a satyr's Mirthful Leaps trait. The second sentence states this correctly. However, it is important to realize that we are considering arbitrary functions. The question of whether Todorcevic's formulation of OCA settles the value of the continuum remains an open problem. has cardinality $\aleph_2$. Are there any infinites not from a powerset of the natural numbers? To learn more, see our tips on writing great answers. 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. 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. Construct a polyhedron from the coordinates of its vertices and calculate the area of each face, Krishna visiting Sudra's home or touching a Sudra. Thanks! @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. Aleph 2, of Cantor's infinite sets X0