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I am a graduate student at UCLA.


1h
comment Showing that a set is a basis of a field as a vector space over a subset of that field
Take an arbitrary element of $F$, express it as an $L$-linear combination of the basis $\{\alpha_1, \ldots, \alpha_m\}$, then expand the coefficients as $K$-linear combinations of $\{\beta_1, \ldots, \beta_n\}$.
15h
answered It's the discrete topology.
1d
comment What is $i^i$?Imaginary, Real, HyperComplex?
@Aditya yes. ${} {}$
2d
comment homomorphism $\phi$ : $\mathbb Z$ $\rightarrow$ $\mathbb Z$ using $\phi(n)$=$2n$
Yes. That it is a homomorphism is called the "distributive property of multiplication over addition". ;-)
Aug
29
comment Are there counter intuitive interpretations of ZF or ZFC?
This is just the standard model of ZF minus the axiom of infinity (i.e., the finite sets), with its elements named in a nonstandard manner.
Aug
27
comment A group with five elements is Abelian
but clearly inappropriate for this particular question.
Aug
26
comment A group with five elements is Abelian
The fact that the size of a conjugacy class divides the order of the group is even less elementary than the fact that the order of a subgroup divides the order of the group.
Aug
25
comment Proving that a function is strictly monotonic knowing that $|f(x)-x^2|\le2|x|$
@AlecTeal Hagan had already addressed my comment in his edit (though his counterexample is discontinuous). Your answer still needs some work.
Aug
25
comment Why is there no natural metric on manifolds?
Different charts will induce different metrics on the same manifold.
Aug
23
comment If $\forall x \in R, x^2-x \in Z(G)$, than $R$ is commutative
"This easily implies $R$ is commutative." Not true at all; consider matrices over a field of characteristic $2$.
Aug
19
comment What makes induction a valid proof technique?
@MauroALLEGRANZA I'd be more inclined to say that in set theory, induction holds by the definition of $\mathbb N$. I typed that before reading Qiaochu Yuan's answer, so I guess I'm saying I'm more in agreement with his way of viewing it.
Aug
19
comment Morphism of schemes $f\colon X\to Y$ associated to a continuous map of the underlying spaces $|X|\to |Y|$
Oh that explains my confusion. Quotient field is just a terrible name to give to the field of fractions.
Aug
19
comment Morphism of schemes $f\colon X\to Y$ associated to a continuous map of the underlying spaces $|X|\to |Y|$
The exponential map $\mathbb R \to \mathbb R$ cannot be a polynomial function, because it grows too fast. In attempting to make this an answer for the first question, you need to define a continuous map $\text{Spec}\mathbb R[x] \to \text{Spec}\mathbb R[x]$ of sets. You've already said what you want the ideal $(x-c)$ to be carried to, and you might as well send the generic point $\{0\}$ to itself. That leaves the remaining maximal ideals that are not of the form $(x-c)$, which exist because $\mathbb R$ is not algebraically closed. Can you work out what the continuity condition means here?
Aug
18
comment Is $\dfrac {dy} {dx} = \dfrac {2x} {3y}$ a homogeneous differential equation?
This differential equation is not linear.
Aug
17
comment Unit circle cannot be well ordered?
Actually, the condition seems to be simply that the topology be at least as fine as the order topology. Nothing more, nothing less.
Aug
17
comment Unit circle cannot be well ordered?
yeah, that's fair, I suppose.
Aug
17
comment Unit circle cannot be well ordered?
I know. I have figured out a proof that intervals are open though; I don't have any reason to believe that you're wrong yet. It just seemed weird to me to give such a convoluted definition, unless it was properly different from the obvious condition that the topology be induced by the order.
Aug
17
comment Unit circle cannot be well ordered?
I understand the solution given in your edit now. I don't understand how you get that this compatibility condition implies the topology is induced by the order, but I guess that's now a separate issue.
Aug
17
comment Unit circle cannot be well ordered?
@AsafKaragila If that were the case, then there would be a much more trivial solution to this problem utilizing the fact that neither topologies have singleton open sets.
Aug
17
comment Unit circle cannot be well ordered?
No, that's not clear to me at all. Everything in $A$ must be less than $y$ and everything in $B$ must be greater than $x$, but, for example, the condition does not require that everything in $A$ be less than everything in $B$.