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Jul
21
comment a question about analysis, how to find the largest cardinality in the following examples
I cannot see how you have resolved the question. Uncountable sets may have larger or if $\neg$ CH smaller cardinality than $|\mathbb R|$, so you haven't shown that B, for instance, is right or wrong.
Jul
20
comment Proving that $\langle \omega+1, \leq \rangle$ and $\langle \omega + \omega^*, \leq \rangle$ are not elementarily equivalent
@Nagase Yes. See what you can do to simplify it now.
Jul
20
comment Proving that $\langle \omega+1, \leq \rangle$ and $\langle \omega + \omega^*, \leq \rangle$ are not elementarily equivalent
@Nagase Even assuming you intended $\le$ in place of $<$, it is incorrect since you have not ensured $z\ne y$.
Jul
13
comment Composite Function Equality
@Jason The a,b nonzero condition should have a negligible effect. I suspect a counterexample still exists, even given those conditions on $g$.
Jul
10
comment Set Theory (Example of Set)
Are you capable of labeling them (with distinct labels) somehow?
Jul
8
revised Composite Function Equality
Polished my answer a little.
Jul
8
comment Composite Function Equality
Yes, +1 for the statement and proof of the (correct) inequality.
Jul
8
comment Composite Function Equality
I don't think you can delete accepted answers. An edit explaining the error is typical, though I am no expert on MSE etiquette.
Jul
8
comment Composite Function Equality
Yes, $f(x)=x$ may have multiple solutions--just have $g$ map all of them to $x^{\ast}$.
Jul
8
answered Composite Function Equality
Jul
8
comment For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic
It becomes obviously right, as the guess the OP made would be a valid isomorphism.
Jul
8
comment For signature $s= \{ f\}, n_f=1$, show that if for structures $M, N$, $f^N, f^M$ have only one orbit, of infinite order, then $M,N$ are isomorphic
@DavidC.Ullrich Well, the first thing OP "didn't tell us" was which way to read an ambiguous, poorly written statement. I suspect the first revision is the problem as stated. There are always going to be conventions/assumptions that appear earlier in the text which an inexperienced mathematician will assume to be standard. The invertibility of a given interpretation of $f$ seems to be one of those, albeit a very odd convention!
Jul
8
revised Injective map between power series ring
added 13 characters in body
Jun
19
answered If a functor $\varphi : C \to C'$ is full, then so is the functor $\varphi \circ$
Jun
16
comment Ring localization and ideals
The set of all elements of $A_S$ of the form $i(a)$ comprise a representation of $A$ in $A_S$. You have, for instance, $\mathbb Z$ in $\mathbb Q$, and of course $\mathbb Q$ has many non-integers. (The example is misleading in that $i$ is injective in that case, which is not always true).
Jun
8
awarded  Strunk & White
Jun
8
revised If $\alpha$ is a root of $f(t) = t^n + a_{n-1}t^{n-1} + \cdots + a_0$, then $|\alpha| \leq n \max_i |a_i|$
added 1 character in body
Jun
5
comment About the Heine-Cantor theorem.
Clearly from the graph...
May
14
comment Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. Then $A'$ is a closed set.
Have you carefully proven that if $A=\mathbb Q$, then no rational number belongs to $A'$? (You're right that the set of irrationals is neither open nor closed)
May
4
comment Simplifying an function with complex numbers
Wait, so you know in advance that both sides are real numbers?