A key idea in topology is you want to capture as much information about a space as necessary, but no more than that. Essentially, a stronger topology has more "information" about a space and a weaker one has less. You want to work with the weakest possible topology that still does what you need.
@AlexBecker I am trying to prove that if $G$ is a free abelian group, then $G$ is torsion free. Would it be helpful to know that $T(G)$, the torsion subgroup is the kernel of the map $f : G \to \Bbb{Q} \otimes_\Bbb{Z} G$ is precisely $T(G)$?
If $G$ is free then $G \cong \oplus_{n \geq 0} \Bbb{Z}^n$
Since tensor products distribute over direct sums and $\Bbb{Q} \otimes_\Bbb{Z} \Bbb{Z} \cong \Bbb{Q}$
This is effectively asking if some associated map $f' : G \to \bigoplus_{ n \geq 0} \Bbb{Q}^n$ is injective
I am now treating $\bigoplus_{n \geq 0} \Bbb{Q}^n$ as a $\Bbb{Z}$ - module.
um....if a free abelian group had a torsion element, it wouldn't be free.
because you would have the relation x^n = 1
you can look at it this way: every subgroup of a free abelian group is free abelian. if x was a torsion element, then <x> is finite, that is: not isomorphic to Z. contradiction.
@OldJohn Do you interpret the sign over the exit door to the pure maths department that read "Beware!! - you are now entering reality" as a statement that pure math is not about reality?
@skullpatrol I'm not sure about that - he maybe didn't intend the book to be a joke! - and probably expected the book to be taken seriously - but I have not read it, so I really can't judge
Ah - but he was a physicist, so probably had a different perspective on reality - compared to a pure mathematician
@JonasTeuwen Not necessarily convergent at all in any mode. Like if the lecturer asks me in the exam for which functions I can compute Fourier series I'm tempted to say any function he can possibly think of if he doesn't want any sort of convergence.
@JonasTeuwen If the lecturer asks me "For which functions can you compute Fourier series" and I say "for any measurable function if you don't make any convergence requirements" I want to know from you how many points from 0 to 10 where 10 is max do I get?
But I need to know for which functions I can compute Fourier series. And now I guess I'd have to say $L^p$ for all $p \geq 1$ and finite measure space if we want the coefficients to be finite but not necessarily want convergence of the Fourier series.
@Matt Then I'll say it now: this is nonsense. If your function isn't integrable then your Fourier coefficients aren't well-defined in the first place. (oscillations may lead to computing $\infty - \infty$.)
I take it he only discussed the Fourier transform on the circle. Then integrable is just fine (as is $L^2$). Both give no problems whatsoever. I don't know what you mean by "finite coefficients". If you have finite coefficients then finiteness of the zero'th coefficient tells you that your function is integrable.
@OldJohn Went up to 41°C, so I called the GP again, he asked a couple of questions and said: well, if it goes up to 41.7°C I will come, but for now it sounds fine!
But cannot get a drivers license for trains with passengers. Need to be 21 for that.
That's the only thing I think.
Well, you can get it. You can go to the train company at 18 and say: "I want to be a train driver!" they will be: "alright! you get an internal education about trains! Cool eh?". So then the next three years you can do small trains which stop like every five minutes 8-).
He doesn't make it explicit, but I understand what we do is define $x\sim x'$ iff $p(x')=p(x)$, and we define $g(b)=G(x)$ for any $x\in \widehat {p^{-1}(\{b\})}$
@Peter: It’s exactly like starting with a group $G$ and a surjective homomorphism $h:G\to H$ and showing that if $K$ is a group, and $f:G\to K$ is a homomorphism such that $f(g)=f(g')$ whenever $h(g)=h(g')$, then there is a homomorphism $g:H\to K$ such that $f=h\circ g$.
@BrianM.Scott i'd say it's "more like" inducing an equivalence relation on a set from a function...there no notion of "multiplicativeness" in topological spaces.
@PeterTamaroff Yes. Specifically, it satisfies $h(x*y)=h(x)\cdot h(y)$, where $*$ is the operation in the domain and $\cdot$ the operation in the range.
@BrianM.Scott OK, so now we define the identification topology: given an onto function $f$from a topological space $X$ to some set $Y$, the topology $\mathfrak I'$ on $Y$ consists of those sets $U\subset Y$ such that $f^{-1}(U)$ is open in $X$. This makes $f$ an identification.
@BrianM.Scott this i know....but as i understand it, Peter's exposure to abstract algebra isn't extensive enough for him to truly appreciate your comparison.
@BrianM.Scott We start with any function $f:(X,\mathfrak I)\to (Y,\mathfrak I')$. We define $x\sim_f x' \iff f(x')=f(x)$. We quotient $X$ by $\sim_f$, and get $X/\sim_f$. $\pi_f :X\to X/\sim_f \;/\;x\mapsto \hat x$ is onto, so we give $X/\sim_f$ it the identification topology determined by $\pi_f$.
@DavidWheeler It is easy to see $f^*$ is one one. Since $f^*$ is continuous, we have that $f^*{}^{-1}(\mathfrak I')\subset \Bbb I$ where the latter is the identification topology, and the former the topology on $Y$.
Now, since $f^*$ is one one, this means $\mathfrak I'\subset f^*(\Bbb I)$
