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Professor emeritus at Cleveland State University. I’m a set-theoretic and general topologist with an interest in combinatorics. I’m also interested in linguistics, especially historical linguistics.


Jun
22
revised Remainders of compactifications are images of the Stone-Čech remainder.
Fixed silly typo.
Jun
22
comment Prove that $\dim(X,\succsim)\leq|X^2|$ - A starting point for a journey into order theory
@Kolmin: You’re very welcome. Yes, it can be very hard to determine the exact dimension of a partial order if it does not have a pretty nice, systematic structure.
Jun
22
comment Remainders of compactifications are images of the Stone-Čech remainder.
@David: You’re welcome. I probably wouldn’t have thought of it either, if I hadn’t known the general result.
Jun
22
answered Remainders of compactifications are images of the Stone-Čech remainder.
Jun
22
answered Understanding Equivalence and Relations
Jun
22
answered Why doesn't sign (appear to) change in inequality?
Jun
22
comment Does $n \log\left(\cos\left(\frac{\pi\,n!}{n^2}\right )\right ) \neq 0 \implies n = p$?
@JohnWO: You’re welcome!
Jun
22
answered Does $n \log\left(\cos\left(\frac{\pi\,n!}{n^2}\right )\right ) \neq 0 \implies n = p$?
Jun
22
comment $k$-cells: Why $a_i < b_i$ instead of $a_i \le b_i$
It allows the $k$ to convey information about the inherent structure of the set: it can be embedded in $\Bbb R^\ell$ if and only if $\ell\ge k$. It’s intrinsically $k$-dimensional.
Jun
22
comment Metacompactness of the Euclidean space
@shir: You’re very welcome.
Jun
21
comment Prove that $\dim(X,\succsim)\leq|X^2|$ - A starting point for a journey into order theory
@Kolmin: I’m having a hard time answering your last question, because I’m not quite sure how you’d apply your algorithm to a more complicated quasiorder. Say you have vertices $1,2,3,a,b,c$. The edges are from $1$ to $b$ and $c$, from $2$ to $a$ and $c$, and from $3$ to $a$ and $b$. (That’s essentially the $n=3$ case of the partial order in my answer.) As you say, if you just fill in the missing pairs arbitrarily, you can easily lose transitivity. // Every partial order is a quasiorder, so if one can get arbitrarily large dimension in partial orders, one can certainly do so in quasiorders.
Jun
21
revised Infix to Postfix
Restored term that I’d inadvertently dropped.
Jun
21
comment Infix to Postfix
@Doug: That’s true of the whole answer: if you check, you’ll see that I inadvertently dropped AND x at the beginning. Fixed.
Jun
21
comment A sequence such that every rational is written infinitely
@user68099: Yes. Call the sequence in which every rational appears infinitely often $\sigma$. For any $x\in\Bbb R$ pick a sequence $\langle q_n:n\in\Bbb N\rangle$ of rationals converging to $x$. Find the first occurrence of of $q_0$ in $\sigma$; that’s the first term of your subsequence; call it $p_{n_0}$. Then find the first occurrence of $q_1$ that occurs after $p_{n_0}$ and call it $p_{n_1}$; that’s the second term of your subsequence. Continue in this fashion, and $\langle p_{n_k}:k\in\Bbb N\rangle$ is a subsequence of $\sigma$ converging to $x$.
Jun
21
comment About $\sum_{n=1}^{\infty}\frac{(-1)^n}{n+(-1)^{n+1}}$
@17SI.34SA: You’re very welcome.
Jun
21
answered About $\sum_{n=1}^{\infty}\frac{(-1)^n}{n+(-1)^{n+1}}$
Jun
21
comment Cantor's Diagonal Argument
Because there is no way to apply it. Exactly how are you going to try to use the argument?
Jun
21
comment How's it possible for each element of the empty set to be even?
@BandeiraGustavo: Basically because it doesn’t fit with the way we want our logic to work. We want the negation of $\forall x(\varphi(x))$, i.e., $\neg\forall x(\varphi(x))$, to be $\exists x(\neg\varphi(x))$. Thus, the negation of every element of the empty set is even ought to be there is an element of the empty set that is not even. The latter statement is certainly false, so we want the former statement to be true.
Jun
21
answered How's it possible for each element of the empty set to be even?
Jun
21
comment A problem on Cauchy sequences
To get real angle brackets for the sequences, use \langle ($\langle$) and \rangle ($\rangle$).