Recent content by Someone2841

One projection will have many 3D representations, so the "vice versa" is wrong.

This post got moved to abstract algebra (I was thinking of ##\mathbb{F}## as a generalization of the real numbers), and so I realize now that I could have just proven that any linearly-ordered group ##\mathbb{G}## that is not Archimedean does not have the least upper bound property. It can be...

Hello! I was wondering if this proof was correct? Thanks in advance! Given: A totally ordered field, ##\mathbb{F}##. Claim: Least Upper Bound Property (l.u.b.) ⇒ Archimedean Principle (AP) --- Proof. I will show that the contrapositive is true; that is, if ##\mathbb{F}## does not have the AP...

The flaw on my reasoning was assuming that ##\lambda *## was continuous from below even with non-measurable sets. This ends up being true: https://math.stackexchange.com/questions/116847/continuity-from-below-for-lebesgue-outer-measure. So no, no such sequence of sets may be constructed.

Sorry if this comes from a place of ignorance, but I notice something and I wonder if you'd comment on it. Since every ##A_n \subset A_{n+1}##, it seems that it follows by induction that ##A_n = \bigcup_{k=0}^n A_k##. If this is the case, it seems that ##\lim_{k\to...

Any ##f:\mathbb{C} \to \mathbb{C}## represents a 2d vector to another 2d vector, so the graph of any such function would be represented by four dimensions.

With regard to the fact that ##v-u>2^{-n}##, your reasoning is sound. It is easy to show that ##2^n>n## for all natural numbers. First, ##2^0=1>0##. Second, ##2^n>n \implies 2^{n+1}>n+1##. Adding 1 to the antecedent, ##2^n+1>n+1##. Since ##2 \cdot 2^n > 2^n+1## is true whenever ##n>0## (verify...

I'm only going to address Q1 and Q2 in this post. Really, since we know the diameter of the nested sequences converges to 0, we also know such a unique ##u## exists. Probably for pedagogical purposes, he shows that there cannot be two elements in the infinite intersection. Because these two...

A cut is not merely a set or pair of sets. Sure, ##\{q \in \mathbb{Q}:q^2<2\}## is a set whose l.u.b. is ##\sqrt{2}##, but it does not meet the criteria for cut. Usually, a cut is an pair of sets A, B that are disjoint and partition ##\mathbb{Q}##; furthermore, A is downward closed and contains...

I think ##F(it)-F(-it)=0##, no? This is unsurprising given that the only occurrence of z in the formula is squared, and ##z^2## is an even function. In any case: ##F(it)=\sqrt{(it)^2+A^2}=\sqrt{-t^2+A^2}## and ##F(-it)=\sqrt{(-it)^2+A^2}=\sqrt{-t^2+A^2}##

Consider the proposed cut you have given and rational numbers p=1 and q=-2. While q<p, p is in the lower set (##x^2<2##) and q is in the upper (##x^2≥2##). Therefore it is not a cut at all, since there are numbers in the upper set (-2) that are less than that of the lower set (1).

I have always thought that non-constant sequences that converge toward 0 in the reals converge toward an infinitesimal in the hyperreals, but recently I have questioned my presumption. If ##(a_n)\to0## in ##R##, wouldn't the same seuqnece converge to 0 in ##*R##? These two statements should...

What about the vector space ##\mathbb{C}=\{(a,b):a,b \in \mathbb{R} \}##, with addition defined as ##(a,b) + (c,d) = (a+c,b+d)##, multiplication defined as ##(a,b)(c,d)=(ac-bd,ad+bc)##, ##\vec 0 = (0,0)##, ##\vec 1 = (1,0)##, and inverses defined by ##-(a,b)=(-a,-b)## (additive) and...

Great point. That totally invalidates my example. :frown: Are there vector spaces that are also fields?

Consider the set ##V = \left \{ f : f \text{is any real-valued function of one real variable}\right \}##. I believe that ##V## is a vector space over the field ##\mathbb{R}##, since for all ##f,g \in V## and ##a,b \in \mathbb{R}##, it is true that ##0 \in V##, ##af-bg \in V##, ##af+ag =...