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1d
answered Help with the proof of the open mapping theorem
1d
comment On Proving that the first euclidean axiom is wrong
The statement that "No line segment can be drawn between that particular point and the point of intersection (C)" seems to be entirely unsupported.
1d
comment On Proving that the first euclidean axiom is wrong
@AsafKaragila Thanks for spotting the typo; I've deleted the comment, and I'll try again.
2d
answered How can Goodstein's theorem be expressed in PA
2d
comment Well-ordered set with greatest element is compact
@AsafKaragila If by "positive proof" you mean a proof by induction on $\beta$ that $]\leftarrow,\beta]$ is compact, then I agree that such a reformulation is easily done, but I'm not sure it's easier to understand --- the essential idea would be unchanged. If, on the other hand, you meant something else, then I'd be interested to know what you had in mind.
Oct
20
answered Well-ordered set with greatest element is compact
Oct
19
answered Is $\mathbb{R}^{[0,1]}$ separable?
Oct
19
comment Why ${1\over n+1}+ \cdot\cdot\cdot +{1\over 2n}>{n\over 2n} $?
There should be an assumption that $n\geq2$, because the strict inequality fails for $n=1$.
Oct
19
comment Why is Skolem normal form equisatisfiable while the second order form equivalent?
As far as I can see, the linked answer is only about the case of second-order sentences obtained by Skolemizing first-order ones, not about general second-order sentences. It's not clear to me how you would even try to go about "de-Skolemizing" something that wasn't obtained by Skolemization in the first place. What if, for example, the function symbol $F$ that you want to get rid of has different arguments at different places in your second-order formula?
Oct
19
comment Why is Skolem normal form equisatisfiable while the second order form equivalent?
Reintroducing the second-order quantifier (or several of them, in more complicated cases) will convert "satisfiable" to "true". Moving the second-order quantifier inside the universal quantifier to return to first order is OK if the formula you're working with originally came from Skolemizing a first-order formula, but in general, starting with an arbitrary second-order existential formula, there will be no way to reduce it to an equivalent first-order formula.
Oct
18
comment Strange thing about Weak Maximum\Minimum Principle?
The intermediate value theorem, which seems to be what you're tacitly using here, has a hypothesis that the domain of the function in question is connected.
Oct
18
answered Why is Skolem normal form equisatisfiable while the second order form equivalent?
Oct
18
answered Strange thing about Weak Maximum\Minimum Principle?
Oct
18
answered Mathematicians' manual of style
Oct
17
answered If I have that all groups $\mid G \mid < 100 \ncong A_5$ are not simple, does this imply solvable?
Oct
17
awarded  Nice Answer
Oct
15
comment Congruences in Algebra
"Unique" in this context means that, even if there are two or more solutions, say $x_1,x_2$, etc., the solutions modulo $n$ (i.e., $x_1\bmod n, x_2\bmod n$, etc.) are all equal so, among solutions modulo $n$, there is only one.
Oct
15
comment If each subset of $L$ with an upper bound has a least upper bound, then each subset with a lower bound a a greatest lower bound
@tmpys The way to deal with this is exactly what I wrote in my previous comment --- one must check that the lub of $B$ is the glb of $S$. That implies, in particular, that the lub of $B$ is in $B$, so that the possibility that you were worried about doesn't actually arise.
Oct
13
answered Can a countable group have uncountably many subgroups?
Oct
12
comment ZF Set Theory and Law of the Excluded Middle
A slight re-pharasing of the previous comments (which are correct) to answer literally the question of which ZF-axioms are used to obtain the law of the excluded middle: No ZF-axioms are used, because the law of the excluded middle is part of the logic on which ZF is based.