if you know what a C* algebra is, presumably here they mean precisely that without the C* identity $\|xx^*\| = \|x\| \|x^*\|$: a Banach algebra with an anti-linear involution.
And a Banach algebra is an algebra on a Banach space with $\|xy\| \leq \|x\|\|y\|$
@MartinSleziak thank you for information, which I will read. I am actually thinking what Mike Miller says and checking some Wolfram Mathworld webpages, like the webpages of C* algebra, Banach algebra, etc.
I'm fairly sure this has been asked before. I'm looking at results along the lines of showing, from first principles (sans Lagrange, etc.), that in a (finite) group, $|\{x\in G| x^3=e\}|$ is odd, or, say, $|\{x\in G| x^5=e\}|$ is a multiple of five.
Let $T:=\{x\in G: x^3=e\}$. Note that (1) $e\in T$ (2) if $x\in T$ then $x\ne x^{-1}$ and finally (3) $x\in T$ implies $x^{-1}\in T$. Hope this helps @Shaun.
user131753
By the way in (2) it should be "if $x\in T\setminus\{e\}$..."
I'm trying to determine the fundamental group of the tensor product sign, regarded as a subset of $\Bbb{R}$, with base point in the middle of the sign. My thought was that the tensor product sign is a deformation retract of the unit disc with four points missing. Is this right? What is the fundamental group of the unit disc with four points missing; how does one compute it?
Your claim about a deformation retract is correct, but it's better to just argue that by collapsing a contractible subset you're left with a wedge of 4 circles.
Wait, which subset is being collapsed? Are you saying that $\otimes$ can be viewed as a wedge product of four circles, which means its fundamental group is the free group with four generators?
Does anyone know any sufficient conditions so that the following diagrams, $\require{AMScd}$ \begin{CD} (X,\tau_X) @>f>> (Y,\tau_Y)\\ @V g V V @AA h A\\ (Z,\tau_Z) @>>\operatorname{id}_Z> (Z,\tau_Z) \end{CD}
and, \begin{CD} (X,\tau_X) @<<f< (Y,\tau_Y)\\ @A g A A @VV h V\\ (Z,\tau_Z) @<<\operatorname{id}_Z< (Z,\tau_Z) \end{CD}
commutes for some $h$ where $(X,\tau_X),(Y,\tau_Y)$ and $(Z,\tau_Z)$ are topological spaces and $f:(X,\tau_X)\to(Y,\tau_Y)$ and $g:(Y,\tau_Y)\to(Z,\tau_Z)$ are given continuous maps?
Collapsing a contractible subset A (satisfying some conditions, eg "NDR") gives a homotopy equivalence X -> X/A
user131753
15:13
Sorry, "..and $f:(X,Ï„_X)→(Y,Ï„_Y)$ and $g:(Y,Ï„_Y)→(Z,Ï„_Z)$ are given continuous maps?" should be "..and $f:(X,Ï„_X)→(Y,Ï„_Y)$ and $g:(Y,Ï„_Y)→(Z,Ï„_Z)$ are given continuous maps in the first case and $f:(Y,Ï„_Y)→(X,Ï„_X)$ and $g:(Z,Ï„_Z)→(Y,Ï„_Y)$ are given continuous maps in the second case?"
Problem: Determine whether $\Bbb{R} / \Bbb{Z}$ is compact...I asked about this question a while ago and I believe I received an answer, but I can't remember it...sorrry...My question is/was, am I to interpret $\Bbb{R} / \Bbb{Z}$ as the a quotient group or a quotient space (i.e., all integers collapse to a single point)?
Consider the set of intersection points from the following curves in the unit square,
$x^s+y^s=1$
$(1-x)^s+y^s=1$
$x^s+(1-y)^s=1$
$(1-x)^s+(1-y)^s=1,$
for $s \in \Bbb Q \ge 1.$
What relation can be formed between all points such that the points form a space? What kind of space is it?
I thi...
I dont see anything wrong with it really. Maybe if somebody complains the issue would need to be revisited but i dont see the point in worrying until then
Consider the set of intersection points from the following curves in the unit square,
$x^S+y^S=1$
$(1-x)^S+y^S=1$
$x^S+(1-y)^S=1$
$(1-x)^S+(1-y)^S=1,$
for $S= \{\Bbb Q \ge 1\}.$
What relation can be formed between all points such that the points form a space? What kind of space is it?
I th...
I think I am going to make a stand: you are not going to learn math effectively (if at all!) if you try to make up your own questions about concepts you haven't really understood from the basics, like the notion of spaces. You absolutely have to get a book - topology, set theory, I don't care - whatever you like - start from the front and get to the back, and do all the exercises.
6
I'm glad to help if you do that. If not, I'm going to stop engaging, I think.
@LeakyNun It is a sum even in the second year of a master's or PhD. Maybe this is not the case if you're studying pure path, where you go into the generalization
@CaptainAmerica16 i'm not really sure but I know that the given is false when the hypothesis is true and the conclusion (Y) is false and X or Y is true when one of them is true
I'm stuck on a problem but I don't want the answer, I want a hint
I'm trying to prove that every Pell equation for a nonsquare $n$ has at least one nontrivial solution.
I've proved that the problem is equivalent to find some rational $r$ satisfying $r^2 - n = 1/m^2$ for some integer $m$. I'm now trying to use analysis/basic topology to find a solution but I can't find how.
Is this true or false?
Let $g:[0,1] \to \Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y \in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wond...
@LeakyNun Haha, perhaps some day. I've recently started my second year (I know you're too iirc but frankly I've always been in awe how you managed to learn all that already).
Consider the set of intersection points from the following curves in the unit square,
$x^s+y^s=1$
$(1-x)^s+y^s=1$
$x^s+(1-y)^s=1$
$(1-x)^s+(1-y)^s=1,$
for $s \in \Bbb Q \ge 1.$
What relation can be formed between all points such that the points form a space? What kind of space is it?
I thi...
Is this true or false?
Let $g:[0,1] \to \Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y \in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wond...
