remeber when I said I get back to you guys after the plane, that's where we went
I cannot do much topology for the following weeks though, cause I am reserving most of the remaining time for quanutm chemistry and theoretical physics
where $S_X$ is an uncountable proper subset picked from $\mathscr{P}(\mathbb{N})$ that is indexed by the interval $X$ (either as a dedekind cut set or an intersection of two dedekind cut sets.
I think that will work. One possible construction is to pick $\mathfrak{c}$ many countable proper subsets of $\mathscr{P}(\mathbb{N})$ indiced by real intervals of the form [x,y], where x,y \in $(i, j]$ with i,j consecutive integers starting from 1. That is
$\{\forall x,y \in (\infty,1]|S_{[x,y]}\cup \{0\}\}$. Then for any two countable subsets indiced by consecutive intervals [i,i+1] and [i+1,i+2], the intersection is guarenteed to be the singleton i+1. If the pair are not consecutive intervals, then the intersection is guarenteed to be the singleton {0}
The intersection obtained by taking the countable set corresponding to the dedekind cut $S_x$ and the dedekind cut (but with the ordered relation reversed) $S_y$
> I wrote two real numbers on two different slips of paper. You choose one at random. You now need to decide whether the one you chose is the larger or the smaller. Can you have a strategy with a correct rate strictly greater than a half?
When you're only working with $\ell^p$ spaces while 95% of textbooks just do things for $L^p$, since the former is covered if you take $\mathbb{N}$ with the counting measure, it's really annoying
im Prinzip wollte ich das jetzt nur bezogen auf die Aufgaben stellung wissen. Ich dachte, $\circ$ ist nur assoziativ, wenn die abbildungen bijektiv sind
nichts. Ich wollte nur verstehen, warum das so gilt. Das, was vorhin gezeigt wurde, hat mir nicht klar gemacht warum eine bijektive abbildung assoziativ sein soll
If $a_1,a_2,\dots,a_n$ are elements of a group $G$, $n\geq 2$, then we define the product $a_1a_2\cdots a_n$ with complete induction to $n$ with \begin{align} a_1a_2&=\text{product of $a_1$ and $a_2$ in the group}& (n=2),\\ a_1a_2\cdots a_n&=(a_1a_2\cdots a_{n-1})\cdot a_n& (n>2). \end{align} For example: $abcde=(((ab)c)d)e)$. With complete induction to $n$ we can deduce easily from the associate law that \begin{align} (a_1a_2\cdots a_k)\cdot(a_{k+1}\cdots a_n)=a_1a_2\cdots a_n&&(1\leq k\leq n-1).
What was the induction hypothesis here? Was it for some $n>2$, we have that for all $k\leq n$, $a_1a_2\cdots a_k=(a_1a_2\cdots a_{k-1})a_k$? Do I have to say the following then: Consider $a_1a_2\cdots a_k$ and $a_{k+1}$. If we multiply these two, we can write $(a_1a_2\cdots a_k)a_{k+1}$. Apply the induction hypothesis, and we're there. Is that's what's happening? Or am I misinterpreting something?
Is anyone willing to help me with a basic question on group theory? It's the last thing I have to understand, and then I've finished my stuff. I posted it here math.stackexchange.com/questions/2155024/… I don't understand why they felt the need to define the order of the brackets, when the operation of a group is associative anyways? So the brackets can move relatively freely anyways?
@SteamyRoot So in the first part they define a sort of 'nested' associativity, and then in the second part they show that associativity works in general?
