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location Cleveland Heights, OH
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visits member for 2 years, 10 months
seen Jan 2 at 12:05

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.


May
27
comment The 'compactness cardinal' of a space
@Paul: You’re welcome!
May
27
answered Big-O: Prove $2^n$ is $O(n!)$
May
27
comment Replacing the value of a function with the value of the limit - is this a standard construction?
@user18921: Ah; if you insist that the limit at $a$ exist for the original function, then indeed this example does not pose a problem.
May
27
comment Replacing the value of a function with the value of the limit - is this a standard construction?
@user18921: Yes, I am. What you called $\text{sgn}^\lambda$ and my two $f^\lambda$s all satisfy your definition, and none of them is maximal.
May
27
answered Replacing the value of a function with the value of the limit - is this a standard construction?
May
27
answered Proof of Real Number Property
May
27
comment Replacing the value of a function with the value of the limit - is this a standard construction?
How are you defining $\lim\limits_{x\to a}f(x)$? Topological spaces needn’t be first countable, so sequences needn’t suffice.
May
27
comment Mergers and ultrafilters
Suppose first that $\mathscr{F}$ is an ultrafilter on $I$, that $X,Y\subseteq I$, and that $X,Y\notin\mathscr{F}$. Then $I\setminus X,I\setminus Y\in\mathscr{F}$, so $$I\setminus(X\cup Y)=(I\setminus X)\cap(I\setminus Y)\in\mathscr{F}\;,$$ and therefore $$X\cup Y=I\setminus(I\setminus(X\cup Y))\notin\mathscr{F}\;.$$ Conversely, if $\mathscr{F}$ is not an ultrafilter, then there is some $X\subseteq I$ such that $X\notin\mathscr{F}$ and $I\setminus X\notin\mathscr{F}$. But then $X\cup(I\setminus X)=I$, and ... ?
May
27
revised What is convex combination of two points?
LaTeX, formatting.
May
27
comment Proving $\sum_{m=0}^M \binom{m+k}{k} = \binom{k+M+1}{k+1}$
@Alan: $\binom{r+n}r=\binom{r+n}n$; now do the translation. (Sorry: I didn’t notice before that you’d used the symmetrically opposite binomial coefficient.)
May
27
revised Create non-substantive filter, belonging chief ultrafilter
added 80 characters in body
May
27
comment Create non-substantive filter, belonging chief ultrafilter
Unfortunately, my previous comment does the opposite of what was requested!
May
27
comment Create non-substantive filter, belonging chief ultrafilter
@Alexander: Thanks, and thanks for the ping.
May
27
answered Create non-substantive filter, belonging chief ultrafilter
May
27
answered The 'compactness cardinal' of a space
May
27
revised How many epsilon numbers $<\omega_1$ are there?
added 211 characters in body
May
27
answered How many epsilon numbers $<\omega_1$ are there?
May
27
comment Create non-substantive filter, belonging chief ultrafilter
Mariya: Let $E=\{2n:n\in\Bbb N\}$, and let $\mathscr{F}=\{F\subseteq\Bbb N:F\supseteq E\}$; clearly $\mathscr{F}$ is a principal filter on $\Bbb N$. Let $\mathscr\{U\}$ be any non-principal ultrafilter on $E$, and let $\mathscr{G}=\{U\cup A:U\in\mathscr{U}\text{ and }A\subseteq\Bbb N\setminus E\}$: $\mathscr{F}\subseteq\mathscr{G}$, and $\mathscr{G}$ is a non-principal ultrafilter on $\Bbb N$.
May
27
comment Create non-substantive filter, belonging chief ultrafilter
This was closed far too hastily. Chief and substantive are pretty clearly mistranslations of some word whose meaning in this context is principal. The question almost certainly asks us to construct a non-principal filter $\mathscr{F}$ that can be extended to a principal ultrafilter $\mathscr{G}$.
May
27
comment Rudin Principles Theorem 2.40: Every k-cell is compact.
@Bryan: I’m not sure exactly which statement you’re trying to prove.