Define ##\Omega: \mathbb{Z_{p^n}} \rightarrow \mathbb{Z_{p^n}}## where ##p## is prime
with ##\ \ \ \ \ \ \Omega(x) = x^{p}##
I am trying to prove this is a ##Hom## under addition.
any ideas?
I tend to agree with you guys. It seems the operation is not really that important though as long as the action on the set is well-defined.
For the sake of discussion, I am including an image of the Author's (A.J Green) definition.
A group ##G## is said to act on a set ##X## when there is a map ##\phi:G×X \rightarrow X## such that the following conditions hold for any element ##x \in X##.
1. ##\phi(e,x)=x## where ##e## is the identity element of ##G##.
2. ##\phi(g,\phi(h,x))=\phi(gh,x) \ \ \forall g,h \in G##.
My...
Maybe I am reading too much into this.
After equation (23), Rudin writes:
"...Assertion (c) would be false if boundedness were omitted from the hypotheses."
Can you explain this further? especially via an example without using the set of integers.
If it was bounded, then ##E## must be closed as well for ##f## to be uniform continuous. I am citing a case where ##E## is
not bounded. Rudin gives the example of ##\mathbb{Z}## and states that ANY function defined on ##\mathbb{Z}## is
indeed uniformly continuous.
##prop:## let set ##E \subset \mathbb{R}## be unbounded, then ##\forall f## well-defined on ##E##, if ##f## is continuous, then ##f## is uniformly continuous.
First am I reading this correctly, and second, I am having a hard time seeing this. Could someone please shed some light on this...
Define:
$$\mathbb{Y} = C \times C^{c} \subset \mathbb{R}^{2}$$
where ##C## is the Cantor set and ##C^{c}## is its complement in ##[0,1]##
First I think ##\mathbb{Y}## is neither open nor closed.
Second, the Hausdorff dimension of ##C## is ##\Large \frac{log2}{log3}##. How do we...
Though this may be related to lin. alg. but it deals with Analysis.
There are 8 axioms for Vector Spaces. To prove a space ##V## is a VS, one must check all 8 axioms (i.e. closure under addition, scalar multi. etc...)
My question is this, it seems cumbersome to have to do this every time...