4,431 reputation
629
bio website math.boisestate.edu
location Boise, ID
age 24
visits member for 1 year, 7 months
seen Jul 22 at 21:45

I am a graduate student in Mathematics at Boise State University.

$\mathsf{I}\; \mathsf{am}\; \mathsf{extremely}\; \mathsf{grateful} $.
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Jul
2
awarded  Curious
May
15
comment Prove that sub-trees have a common vertex
@NicholasR.Peterson Oh whoops I was reading $\ne$ as $=$
May
13
comment Example of metrics that generating the same topology but not uniformly equivalent
Nice, good example.
May
9
accepted Showing $2$ is not definable in $(\mathbb{Q},+)$.
May
9
comment Showing $2$ is not definable in $(\mathbb{Q},+)$.
Yeah so I wrote that down. Assume $2$ is definable, that is $\mathcal{A}=\{2\}$ is a definable set. All one needs is an automorphism, (say $x\mapsto 2x$) that does not fix $2$. We have that $2\in \mathcal{A}$ if and only if $2+2 \in \mathcal{A}$. Obviously the automorphism $x\mapsto 2x$ does not fix $2$, but I was thinking that this argument was too simple. I guess I'll have to look at the invariance of automorphisms proof once more.
May
9
accepted Ramsey Theory: Showing the existence of a special set of natural numbers.
May
9
comment Showing $2$ is not definable in $(\mathbb{Q},+)$.
@AsafKaragila Yeah, it is. I already turned my portfolio in though. Now I'm just working on problems for fun.
May
9
asked Showing $2$ is not definable in $(\mathbb{Q},+)$.
May
7
comment $\mathsf{ZF} \vdash \exists x \forall y (y\notin x)$
@MauroALLEGRANZA Why do you think the axiom of separation is: $\exists x \forall y ( y\in x \longleftrightarrow x \in a \land \phi(y) )$--I've never seen it written like that before. I write it as: $\exists x \forall y ( y\in x \longleftrightarrow \textbf{y} \in a \land \phi(y) )$
May
7
comment Ramsey Theory: Showing the existence of a special set of natural numbers.
I guess what I've proved is sort of interesting. Under this coloring, either there is a monochromatic set such that the sum of every two numbers of that set has an even number of prime factors or there is a monochromatic set such that each prime $p$ divides all but finitely many of its elements.
May
7
revised Ramsey Theory: Showing the existence of a special set of natural numbers.
added 178 characters in body
May
7
comment Ramsey Theory: Showing the existence of a special set of natural numbers.
@AndresCaicedo I just realized I could have used the set: $\{pn+1: n \in \omega\}$ for fixed prime $p$, which makes the problem a triviality.... Also, I realized, the false logic there too
May
7
answered Ramsey Theory: Showing the existence of a special set of natural numbers.
May
7
revised Ramsey Theory: Showing the existence of a special set of natural numbers.
edited tags
May
7
asked Ramsey Theory: Showing the existence of a special set of natural numbers.
May
5
comment $\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.
Samuel Coskey, you know him?
May
5
accepted $\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.
May
5
comment $\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.
I see, elementary equivalence will not do the trick. We need completeness which is not first-order expressible. But it is second-order. We haven't got to second-order logic in our course yet.
May
5
revised $\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.
added 295 characters in body
May
5
comment $\mathbb{R}$ and $\mathbb{R}\setminus \{0\}$ are not isomorphic as linear orders.
Hey thanks, much appreciated. I should have realized this-- ARG