1,549 reputation
511
bio website
location
age
visits member for 3 years, 7 months
seen Jul 1 at 18:17

Having a bit o' fun with math.


Feb
11
suggested suggested edit on Cardinal arithmetic
Feb
11
answered What is an efficient algorithm to compute modular exponentiation of stacked exponents?
Feb
11
answered Fastest way to compute HCF of 2 numbers
Feb
10
comment Cardinality of all lines on $\mathbb R^{2}$ which do not contain point $(x,y)\in l$ where $x, y \in \mathbb Q$
How many irrationals are there? (If there are $\aleph_0$ rationals and $2^{\aleph_0}$ total real numbers, then ...) There is one horizontal line for each irrational. This will give you a lower bound (since there are lots of non-horizontal lines that miss rational points) on the number of lines you are looking for.
Feb
10
comment Stone-Cech compactification problem
Yeah, and for general spaces, i guess the same argument should work (if one is okay with non-Hausdorff compactification.)
Feb
10
answered Injective maps from $B^A$ to $(C^B)^{C^A}$
Feb
10
answered Cardinality of all lines on $\mathbb R^{2}$ which do not contain point $(x,y)\in l$ where $x, y \in \mathbb Q$
Feb
10
answered Stone-Cech compactification problem
Feb
9
awarded  Revival
Feb
9
answered Alexandroff compactification question
Feb
9
awarded  Mortarboard
Feb
8
revised Cardinality of $P(\mathbb{R})$ and $P(P(\mathbb{R}))$
added 126 characters in body
Feb
8
answered Cardinality of $P(\mathbb{R})$ and $P(P(\mathbb{R}))$
Feb
8
awarded  Necromancer
Feb
8
comment Size of the closure of a set
Yes, I wouldn't be surprised if there weren't an independence result here, something like $\text{MA}+2^{\aleph_0}=\aleph_2$ implying that every separable, sequentially-compact, (+compact?) space is $\leq\aleph_2$. For example, there's the result that if $\mathfrak{c}\leq\aleph_2$ then every compact, sequentially-compact space is pseudo-radial, etc.
Feb
8
comment Size of the closure of a set
One more comment: the claim at that "For pseudoradial Hausdorff spaces X we have |X| <= d(X)^c(X) (<= 2^d(X))" (at the at.yorku.ca link) is not correct (it is the second inequality that is a problem.) Otherwise we would have a fine contradiction coming from the fact that if $\mathfrak{c}\leq\aleph_2$ then all compact, sequentially compact spaces are pseudo-radial, but my (consistent) counterexample is a separable, compact, sequentially-compact space (hence pseudo-radial) and $\mathfrak{c}=\aleph_2$ but $|X|>2^{\aleph_0}$.
Feb
8
comment Size of the closure of a set
This is the Hewitt-Marczewski-Pondiczery Theorem: let $\kappa\geq\aleph_0$ be a cardinal and $\beta\leq2^\kappa$. Then if $X_\alpha$ are topological spaces with $d(X_\alpha)\leq\kappa$ for $\alpha<\beta$ then $d(\prod_{\alpha<\beta}X_\alpha)\leq\kappa$. In our case, we are using the special case for $\kappa=\aleph_0$: the product of no more than continuum many separable spaces is separable. We are taking the product of $\omega_1<2^\omega$ many finite spaces. ($d(X)$ is the density of a space $X$.)
Feb
8
comment Size of the closure of a set
Yes. It's a great book if you're studying set-theoretic topology.
Feb
8
awarded  Editor
Feb
8
comment Size of the closure of a set
Thanks, I stuck in the forcing construction, since it's not a priori obvious how to force such a model.