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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.


Aug
8
comment a question on delta systems
@ctlaltdefeat: Yes, that's right.
Aug
8
comment Obtain a grammar for the language (i) L = {a ^m b ^n ; m ≠ n ; m , n > 0 }
The first paragraph really ought to be a comment.
Aug
8
comment How do I determine if the set of triples of all real numbers on $\mathbb{R}$ is a vector space.
Your question is not actually off-topic in the usual sense of the term: there's nothing wrong with the topic at all. Some people here are venomously opposed to questions that do not include some of your own work and seem to go out of their way to put such questions on hold. You can protect yourself from them by including some indication of what you've tried. Fortunately, you got some useful answers first.
Aug
8
answered Isolated points in $B(S)$?
Aug
8
answered Ordering friends
Aug
8
comment a question on delta systems
@ctlaltdefeat: $\beta_0<\kappa$, so $|\beta_0|<\kappa$, i.e., $|\beta_0|\le 2^\omega$. Consider the sets $S_\gamma\cap\beta_0$. Each of them is countable; how many countable subsets of $\beta_0$ are there? How many sets $S_\gamma$ are there?
Aug
8
comment if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or…
... open nbhd $U$ of $z_k$.
Aug
8
comment if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or…
@fatemeh: As I said, it is not necessarily possible to enumerate $D^d$ as $\{x_n:n\in\omega\}$, unless you allow a point of $D^d$ to be enumerated infinitely often if $D^d$ is finite. The proof that you have slides over this small difficulty; I didn't. $\alpha=|D^d|$. If $D^d$ is infinite, this is $\omega$; if not, it's some finite $n<\omega$. // As long as $z_k$ has an open nbhd $U_k$ such that $D_k\setminus U_k$ is infinite, we define $z_{k+1}$ and $D_{k+1}$, and the construction continues for another step. It stops only if we get some $z_k$ such that $D_k\setminus U$ is finite for every ...
Aug
8
comment $ KC $ spaces imply $ US $ spaces , but vise versa is false.
@fatemeh: I'm sorry: $D$ was a typo for $F$. I've fixed it now. $F$ is nowhere dense because a convergent sequence in $[0,1]$ together with its limit point is a closed set that does not contain any open interval.
Aug
8
revised $ KC $ spaces imply $ US $ spaces , but vise versa is false.
Typo.
Aug
7
awarded  Nice Answer
Aug
7
comment Choosing $n$ objects in $2^{2n}$ ways
@CommanderShepard: You're welcome.
Aug
7
answered Choosing $n$ objects in $2^{2n}$ ways
Aug
7
comment Ted Sider's Definition of a Total Relation over a Set D
It is not a mistake. The relation $\mathscr{D}\times\mathscr{D}$ is sometimes called the total relation on $\mathscr{D}$; I use the term myself.
Aug
7
comment Assume that the probability is $\frac12$ that a child born is a boy
Putting this question on hold is a travesty, if not an outright abuse of power: the OP has obviously done some work, since the question included tentative answers. Moreover, the answers are right, which is evidence that the OP has some understanding.
Aug
7
answered if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or…
Aug
7
comment if $ X$ is a countable, compact $ T_{1} $ space and $ A ‎\subseteq‎‎‎‎ X $ then either $A$ is compact or…
@savick01: The derived set of $A$.
Aug
7
answered Recurrence Relation Homework Question 3
Aug
7
answered $ KC $ spaces imply $ US $ spaces , but vise versa is false.
Aug
7
comment An elementary question about images of relations.
@Jose: You're welcome.