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Mar
31
comment Let $|f(z)| \to \infty$ as $|z| \to \infty$, prove that $f(\mathbb{C})= \mathbb{C}$?
@Chappers Continuity gives you this.
Mar
31
answered Let $|f(z)| \to \infty$ as $|z| \to \infty$, prove that $f(\mathbb{C})= \mathbb{C}$?
Mar
31
comment Prove that $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$
@user123733 Draw a picture of a sequence $k_n$, hopefully this will make things clearer.
Mar
31
comment Prove that $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$
Just take $\epsilon=\frac{b-a}2$. The reasoning above shows that there exists some $i$ with $k_n\in(a,b)$.
Mar
31
comment Prove that $\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}$ is dense in $[0,1]$
@user123733 The key point is that the $1-\epsilon\leq k_0\leq 1$, and for all $i$, $0<k_{i}-k_{i+1}\leq \epsilon$ and $k_i\to 0$. This means that any $c$ is $\epsilon$-close to some $\phi(n)/n$
Mar
30
comment Example of $H^n(X,R)$ not equal to $Hom(H_n(X,R),R)$
@Dan You seem to suggest that my question was designed merely as a way to show off, when in reality I was trying to rule out either a confusion or a typo: the universal coefficient theorem, at least the one I know, relates homology with coefficients in a PID to cohomology with coefficients in a module over said PID. How is suggesting that there may be a $\Bbb Z$ (or any other PID for that matter) rather than a $G$ for the coefficients of homology a rude question when the question doesn't seem to make sense without it?
Mar
30
comment Example of $H^n(X,R)$ not equal to $Hom(H_n(X,R),R)$
Why are there two $G$'s in the $Hom$ both in the title and the body of the question)? When you replace the first $G$ with $\Bbb Z$ in the title, then this is an isomorphism, because $H_0$ is free abelian group, and thus the $Ext^1$ vanishes.
Mar
30
answered How do I show $SO(n)$ is open and closed in $O(n)$?
Mar
27
revised Elementary proof that $\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}$ is conditionally convergent.
This question isn't about complex analysis.
Mar
27
comment Vector bundle of dimension $\leqslant n$ on $n$-connected space is trivial
You can have an $n$-skeleton equal to a point, not just the $(n-1)$-skeleton.
Mar
27
answered If R is a PID, is it true that $R/\ker \phi$ is also a PID?
Mar
26
awarded  Popular Question
Mar
26
reviewed Close Finding the volume of a solid of revolution
Mar
26
reviewed Close show that $X$ is compact
Mar
26
comment Does $\sin(n)$ hit all real numbers between $-1$ and $1$ if $n$ is an integer?
@DanielEscudero It's equivalent to the irrationality of $\pi$.
Mar
25
revised Every nilpotent left ideal is contained in a nilpotent 2 sided ideal.
added 1 character in body
Mar
25
comment Every nilpotent left ideal is contained in a nilpotent 2 sided ideal.
@user225311 yes, fixed it.
Mar
24
comment $\int_{\mathbb R}|f(x)|^{2} dx <\infty \implies \sum_{m\in \mathbb Z}\int_{m-\beta}^{m+\beta}|f(x)|^{2} dx <\infty$?
Yes, because you will cover every point in $\Bbb R$ a set amount of times, smaller than some constant that is roughly equal to $2\beta$ I would guess. The total sum should be smaller than $k\int|f|^2$ for some integer $k$ of the order of $2\beta$.
Mar
23
comment The closed unit ball is not compact in infinite dimension spaces. Why?
Their distance in the sup norm is $1$.
Mar
23
comment Reducibility of $P(X^2)$
So it turns out that one can recast a vote if the answer has been edited recently, which is why I added an invisible string of characters at the end of yours, and was finally able to undo my erroneous downvote.