Show that $f:(0,1)\rightarrow \mathbb{R}$ given by $f(x)=\frac{1}{x}$ is not uniformly continuous.
proof: Let $\epsilon=1$ let $\delta>0$ arbitrary. Set $x\in (0,1)$ to be such that $x<\frac{\delta}{1+\delta}$ and $y= x+\delta$ . Then $|\frac{1}{x}-\frac{1}{x+\delta}|$ $=$ $|\frac{\delta}{x(x+\delta)}|$ $\geq$ $\frac{\delta}{x(1+\delta)}>1$
@TedShifrin suppose the real numbers are enumerated $c_0, c_1, \cdots$. Then let $n_0$ be the smallest $n$ such that $c_n > c_0$. Then $n_1$ be the smallest $n$ such that $c_0 < c_n < c_{n_0}$, etc and then the limit cannot be one of $c_n$
Having trouble proving that given $2^{n+1}$ integers, we can find $2^n$ whose sum is divisible by $2^n$. I'm trying induction. So I assume for IH that it holds true for $2^{k+1}$ integers. Then for IS I have $2^{k+2}$ numbers which can be broken into two groups of size $2^{k+1}$, letting us apply our IH on them. However, I'm not sure how to proceed in combining these results to conclude that with $2^{k+2}$ numbers, there exist $2^{k+1}$ whose sum are div. by $2^{k+1}$
OK, so if you've checked that, I can stipulate that I only care about positive $\delta<1/2$, and then you can check that $y$ will be in the interval, too.
Yes. So if I set epsilon $=1$, and assume $\delta$ is between 0 and 1\2 and $|x-y|<\delta$ and $|f(x)-f(y)|\geq \1$ then it holds for $\delta \geq \frac{1}{2}$
Having trouble proving that given $2^{n+1}$ integers, we can find $2^n$ whose sum is divisible by $2^n$. I'm trying induction. So I assume for IH that it holds true for $2^{k+1}$ integers. Then for IS I have $2^{k+2}$ numbers which can be broken into two groups of size $2^{k+1}$, letting us apply our IH on them. However, I'm not sure how to proceed in combining these results to conclude that with $2^{k+2}$ numbers, there exist $2^{k+1}$ whose sum are div. by $2^{k+1}$
Lol that's literally the question though...I already know it holds true and ive done cases to see that but Idk how to prove it. I showed my work. What am I missing? What should I change...? @LeakyNun
"There is nothing wrong with abstraction and generality -- they are still cornerstones of the mathematical enterprise. But "abstract" is a verb as well as an adjective: general ideas should be abstracted from something, not conjured from thin air. Abstraction in this sense is highly non-Bourbakiste, best summed up by the counter-slogan "let $2 = n$."
To do that we have to start with case 2, and fight our way through it using anything that comes to hand, however clumsy, before refining our methods into an elegant but ethereal technique which -- without such preparation -- let us prove case $n$ without having any idea of what the proof does, how it works, or where it came from." -- Ian Stewart, Galois Theory
I gave the question, I showed my thought process and work leading up to it and what I was stumped on. What am I doing wrong / how do I proceed? @LeakyNun
Like I just went to this puzzle-writing meeting, and I gave this puzzle I'd been thinking about for a few months and this other puzzle I came up with years ago and got reminded of
Serre linear representations of finite groups, page $163$: Let $A$ be a ring, and let $\mathcal{F}$ be a category of left $A$-modules. The Grothendieck group of $\mathcal{F}$, denoted $K(\mathcal{F})$, is the abelian group defined by generators and relations as follows: Generators. A generator $[E]$ is associated with each $E\in\mathcal{F}$. Relations. The relation $[E]=[E']+[E'']$ is associated with each exact sequence: $$0\to E\to E'\to E''\to 0,$$ where $E,E',E"'\in\mathcal{F}$.
just to check, there is an error in the above right this isn't equivalent to taking the relation as $[E']=[E]…
The $n\times n$ matrix A whose columns are the coordinate vectors of the images $F(\mathbf e_1),...,F(\mathbf e_n)$ with respect to the basis $\mathbf e_1,...,\mathbf e_n$ is called the matrix of $F$ with respect to the basis $\mathbf e_1,...,\mathbf e_n$. (1)
Now consider $\mathbf x =T\mathbf x'$. $T$ is the matrix whose columns are the coordinate vectors of $\mathbf e_1',...,\mathbf e_n'$ with respect to the basis $\mathbf e_1,...,\mathbf e_n$.
Does $T$ respect the definition (1)? Isn't $T$ the matrix of a linear transformation that takes a vector $x_1'\mathbf e_1'+x_2'\mathbf e_2',...,x_n'\mathbf e_n'$ and returns $x_1\mathbf e_1'+x_2\mathbf e_2',...,x_n\mathbf e_n'$, and so the columns of $T$ should be the coordinate vectors with respect to the basis $\mathbf e_1',...,\mathbf e_n'$.
OK, so, since (by the way it's PHI) $\phi(v_1), \dots, \phi(v_n)$ are linearly independent, what do you conclude if $c_1\phi(v_1) + \dots + c_n\phi(v_n) = 0$?
Yup. 99% of the time, if you want to prove some vectors are linearly independent, you automatically write down the "Suppose ..." sentence as I did (with those vectors in there).
If{v1,...,vn}is linearly independent, then{φ(v1),...,φ(vn)}is linearly independent. so fo this is it necessary that linear map to be injective if yes can you give me some counter example?
Oh, it seems the name is ambiguous. I'm talking about the identity ${\alpha+\beta\choose n}=\sum_{k=0}^n{\alpha\choose k}{\beta\choose n-k}$ with $\alpha,\beta\in\mathbb{C}$. The presence of complex numbers sadly rules out the neat combinatorial arguments. Thanks regardless.
I'm doing this as part of a second-week exercise to a complex analysis lecture. The machinery of contour integration isn't at my disposal yet. If there isn't a nice algebraic proof, I should probably change my approach altogether. Thanks for the input.
If $\mathfrak{g}$ is a Lie algebra, then the killing form on it is $K(x,y) = tr(ad~x ~ ad~y)$. I'm thinking about the case when $\mathfrak{g} = \mathfrak{sl}_n(\Bbb{C})$. But, e.g., $ad ~ x$ is not a matrix, so how do I compute the trace of $ad ~x ~ ad ~y$?
