Someone practiced in homological algebra, tell me if I'm wrong: if I have a contracting homotopy on a chain complex $C_\bullet$, then won't I get a contracting homotopy on the cochain complex $\textrm{Hom}(C_\bullet, A)$?
@anon I'm sure you realised this already, but: A linear G-action is a map $G \times V \to V$, while a homomorphism from G to GL(V) is a map $G \to (V \to V)$ (using some idiosyncratic notation). They are not the same, but they aren't that different either... However, I agree an introductory text ought to explain the connection.
Our analysis lecturer said in class yesterday by the contraction mapping principle the function $f(x) = x$ defined on any interval $I$ of $\Bbb{R}$ has a unique fixed point
Don't we need that the interval $I$ be closed (and hence complete) in order for the contraction mapping principle to apply?
@Srivatsan Can you send me that link again? It was the link on Arturo's answer talking about how in mathematics, we have like the intersection of all closed sets containing a set A to be the smallest closed set containing it
@MattN Well I know the definition. But why do we introduce them?
We want to study properties of subsets $A$ of an ambient group $Z$. Now if $A$ is contained in a subgroup of $Z$ then some nice identities hold. For example, if $A$ is itself a subgroup of $Z$ we get $$ \widehat{\chi_A} = \textbf{P}_Z(A) \chi_{A^\bot}$$ Unfortunately, not all ambient groups have subgroups (take for example $Z = Z_p$ for $p$ prime). In this case we can "save" ourselves and still get some nice identities by introducing Bohr sets which in some sense are a generalisation of subgroups of $Z$.
@Daniil If I deemed Ilya my friend I would flag that message as offensive, "because it seemed to me to be offensive to" Ilya ;-) but my name is not Fortuon...
In this answer by mixedmath, he says "For a nontrivial sum of consecutive natural numbers to be even, there must an an odd number of terms." But of course 13 + 14 + 15 + 16 + 17 + 18 is even, so unless he's using a different definition of "nontrivial"...
@robjohn (the 2nd) could you please take care of this? It is exactly the same question by the same OP posted within a few hours, hence nothing will be lost by a deletion.
So if we could figure out the distances to the nearest grocery store for both anon and robjohn we could estimate average trip times. Then, when that robjohn comes back, we might be able to guess which one he is based on how long he was gone.
@Chris: You should try thinking about it. I knew you were trying to write 2^{f(n)} (you need curly brackets to put things in exponents), not 2*f(n). If you think what I wrote is a counterexample to if f(n)=o(g(n)) then 2f(n)=o(2g(n)) then I'm not sure you're even trying.
@robjohn: Thank you for pointing this out to me (about the curly brackets). Yes, you're right, I am a bit off now, as I've been studying for several hours. I've done this:
$ 1/n < c*2 => 2^{1/2}/2 < c => 2^{-1/2} < c => 1/SquareRoot(2) < c $ Which for $c=1$ is not true. So the statement is false.
Where can I find these functions? I didn't get this " \implies for logical implication ". We have been given the 1st assignment on algorithms and complexity course and there was a question in which we had to prove if the statements where false/true.
