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visits member for 2 years, 6 months
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Oct
15
comment Completion of rational numbers via Cauchy sequences
What do you mean "maybe"? Is this property somehow equivalent to transitivity as I stated above?
Oct
15
comment Completion of rational numbers via Cauchy sequences
Why is Definition 1 property 1 called transitivity? Doesn't transitivity mean that $\forall x,y,z \in K : (x \leq y \land y \leq z) \implies x \leq z$?
Sep
24
awarded  Autobiographer
Aug
26
revised Every DPDA has an equivalent DPDA that always reads the entire input string
deleted 224 characters in body
Aug
25
awarded  Informed
Aug
24
revised Every DPDA has an equivalent DPDA that always reads the entire input string
added 14 characters in body
Aug
24
revised Every DPDA has an equivalent DPDA that always reads the entire input string
edited tags
Aug
24
revised Every DPDA has an equivalent DPDA that always reads the entire input string
edited tags
Aug
23
asked Every DPDA has an equivalent DPDA that always reads the entire input string
Aug
19
awarded  Tumbleweed
Feb
4
comment Russell's Paradox in *Naive Set Theory* by Paul Halmos
@GME: that helps me understand somewhat. However, I still can't see how we know that we can't assume $B \notin A$ and then contrive some other contradiction. Help?
Jan
26
revised Russell's Paradox in *Naive Set Theory* by Paul Halmos
deleted 36 characters in body
Jan
26
asked Russell's Paradox in *Naive Set Theory* by Paul Halmos
Jan
9
revised Can we always construct an element not on a list?
added 97 characters in body
Jan
9
asked Can we always construct an element not on a list?
Jan
7
comment The language of abstract algebra in $ab=a, ab=b$ implies $a=b$
I have no trouble applying this principle to objects which are somehow explicitly constructed. However, in this case, $a$, $b$, and $ab$ do not seem to me to be explicitly constructed and so I have trouble applying the principle.
Jan
7
comment The language of abstract algebra in $ab=a, ab=b$ implies $a=b$
So you are referring to the ambiguity between a function and the category theoretic notion of a morphism? I haven't studied category theory yet, but maybe it would help me with this problem?
Jan
3
comment The language of abstract algebra in $ab=a, ab=b$ implies $a=b$
I was referring to the binary operation of $G$. I'm not sure what you are referring to.
Jan
2
comment The language of abstract algebra in $ab=a, ab=b$ implies $a=b$
What is ambiguous about the term "map"?
Jan
2
comment The language of abstract algebra in $ab=a, ab=b$ implies $a=b$
@user44197 Typo fixed.