Recent content by Bachelier

yes, even in the general case ##\Omega(1+1) = 2^p \neq 2 = \Omega(1) + \Omega(1) \ \forall n, p \geq 2##

You did notice that the ring I am working with has ##p^n## elements. It is an extension field of ##\mathbb{Z_{p^n}}##.

The ring ##K## has order ##p^n## which makes it ##\cong \mathbb{Z_{p^n}}##

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.

Yea I agree. That is why I am asking. Please see the attached theorem 4.20. The assumption on the boundedness is at the end of page 2

So that gives the ##Hausdorff \ dim## of ##C^c## to be 1. makes sense.

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...

Cool. Thank you.

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...

Interesting 1,2,4,8 Theorem. Thank you for sharing.