@AlessandroCodenotti I realized this really boils down to local behaviour of dimension like you were trying to do. We have that the local small inductive dimension and small inductive dimension is the same, so surjective local homeomorphism preserves $\text{ind}$
**Theorem (First Existence Theorem of Intermediate Values):** A continuous function defined on a closed interval $[a, b]$ takes on all values between $f(a)$ and $f(b)$.
**Proof:** Without loss of generality, assume $f(a) \leq f(b)$. The objective is to prove that for any $y_0 \in [f(a), f(b)]$, there exists $x_0 \in [a, b]$ such that $f(x_0) = y_0$.
If $y_0 = f(a)$, then simply let $x_0 = a$. Similarly, if $y_0 = f(b)$, then let $x_0 = b$. Now, consider the case where $y_0 \in (f(a), f(b))$. Define the function:
More formally: If $f:X\to Y$ is such local homeomorphism, then cover $X$ by open sets for which restriction of $f$ to them is a homeomorphism onto its (open) image. The supremum over small inductive dimension of those sets is $\text{ind}(X)$, and their images have the same dimension. The supremum small inductive dimensions of their images is $\text{ind}(Y)$.
In partition of unity, locally finite is assumed so that the final sum $\sum_i \theta_i(x) = 1$ is a finite sum . Why do we need this sum to be finite? Why is it a problem if the sum is not finite? Don't we only care about the sum equaling one?
**Theorem (First Existence Theorem of Intermediate Values):** A continuous function defined on a closed interval $[a, b]$ takes on all values between $f(a)$ and $f(b)$.
**Proof:** Without loss of generality, assume $f(a) \leq f(b)$. The objective is to prove that for any $y_0 \in [f(a), f(b)]$, there exists $x_0 \in [a, b]$ such that $f(x_0) = y_0$.
If $y_0 = f(a)$, then simply let $x_0 = a$. Similarly, if $y_0 = f(b)$, then let $x_0 = b$. Now, consider the case where $y_0 \in (f(a), f(b))$. Define the function:
@Jakobian there's actually a cute fact: any family of $[0,1]$-valued functions that is point-wise summable determines an open cover by preimages of $(0,1]$ on which a partition of unity exists
also, if that helps your conscience, as my reply to Jakobian above indicates, there is no sensible way of generalizing the notion to yield anything more general
@psie here's a naïve try: $\sum_{n\in\mathbb Z} c_n$ is just the function evaluated at $0$, so we have $$1=\sum_{n\in\mathbb Z} c_{2n}+\sum_{n\in\mathbb Z}c_{2n+1}=\frac{2}{2\pi}\int_{-\infty}^\infty \frac{\sin x} x dx.$$ And by evenness, we conclude that $\int_0^\infty \frac{\sin x} xdx=\frac{\pi}{2}$. But for this to work, I guess one requires absolute convergence of $\sum c_n$, so that one can rearrange it.
@Sahaj They are both showing signs of age. Trump has the advantage of having bullshit for so long that he can do it even while going senile. My great uncle, who had a great gift of gab, could use that after he had severe Alzheimers to make it seem as if he were fine. It was only when he did not recognize me that I knew he was not all there.
**Theorem (First Existence Theorem of Intermediate Values):** A continuous function defined on a closed interval $[a, b]$ takes on all values between $f(a)$ and $f(b)$.
**Proof:** Without loss of generality, assume $f(a) \leq f(b)$. The objective is to prove that for any $y_0 \in [f(a), f(b)]$, there exists $x_0 \in [a, b]$ such that $f(x_0) = y_0$.
If $y_0 = f(a)$, then simply let $x_0 = a$. Similarly, if $y_0 = f(b)$, then let $x_0 = b$. Now, consider the case where $y_0 \in (f(a), f(b))$. Define the function:
why with a function other than f(x) , is the theorem proven?
it should go without saying that some people use dark mode because they prefer it, not because of any problems with their eyes, or beliefs about how 'good' it is for them
@leslietownes the book prove the final part of the theorem with a function g(x) , which is different from f(x) , and I don't understand why this is valid
so if you're looking to prove that a solution to f(x) = y_0 exists, and you have a result that you can use like a black box to prove that roots of functions exist, you can use that result in any way you like
including by applying it to a function that isn't named "f"
a background thing that's maybe confusing here, is why this textbook/treatment has chosen to give the "assumes the value 0" case of the intermediate value theorem a prominence that it maybe doesn't deserve
but we don't need to know why they did that, to be able to use that case like a black box
I know that if you change from dark to white mode, it can feel bad, even as it "burns your eyes". And I think this is just that, you don't have developed light receptors because you were using too much dark mode. Similarly pirates used this as a strategy, that's why they wear eye-patches
To have well-developed receptors you can't sit only in the dark, otherwise you'll feel like the light is burning your eyes
I kind of like the largest splitting topology on $C(X, Y)$, its like this esoteric topology that you don't know how to write explicitly unless $X$ is locally compact, where it coincides with compact-open
Thinking about it, this should be your go-to topology on $C(X, Y)$ probably
i.e. one for which continuity of a map $T\times X\to Y$ implies continuity of $T\to C(X, Y)$
perhaps that would be better thing to do than considering compact-open
Basic question. I know that under absolute convergence $$\sum a_n=\sum a_{2n}+\sum a_{2n+1}.$$ Can the "absolute convergence" condition be relaxed a bit so that it isn't required, but something else?
psie i don't know what is originating this question but i suspect you might be thinking way too formally about all of this. has anyone in your book or class or whatever it is said 'just do this formally'
i say this only because fourier series (where i think this is coming from because of your general questions) are notorious for being difficult to analyze and not being amenable to the usual calc 2 convergence tests even when they converge
if someone is writing "sum_{n in Z} a_n" without saying anything about how they are ordering Z or otherwise exhausting Z by finite subsets, and without any guarantee that the sum is absolutely convergent, they are probably inviting you not to care about those things
@Thorgott what a funny coincidence, I got confused by the difference between the compact open topology and convergence of the images in the Vietoris topology earlier today
Namely, consider the lattice of open sets in $[0, 1]$ modulo the equivalence relation that you identify two open sets if they differ by measure 0 sets.
This is called a "locale" or a "pointless topology"
The "pointless space" we create in this way is the "locale of random reals between 0 and 1"
oh I know. What if we take topology generated by sets $\{f : f(A)\subseteq U\}$ where $A$ is closed and $U$ is open and try to show its splitting?
it won't be splitting in general but what about $\mathbb{Q}$
or instead of all closed sets just take some closed sets
then we could apply lemma 2 from Fox paper and obtain that $h\mapsto h\cdot f$ is not continuous or whatever, maybe he meant $h\mapsto f\cdot h$ I don't know
we'd have to show that continuity of a map $T\times\mathbb{Q}\to \mathbb{R}$ implies continuity of an induced map $T\to C(\mathbb{Q}, \mathbb{R})$
this should be equivalent to pre-image of $M(A, U) = \{f: f(A)\subseteq U\}$ being open in $T$
(say, we only add it for one closed non-compact set $A$ for simplicity)
If $g:T\times\mathbb{Q}\to\mathbb{R}$ denotes this map then the other map is given by $\hat g:t\mapsto g(t, \cdot)$
$\hat g^{-1}(M(A, U)) = \{t\in T : g(\{t\}\times A)\subseteq U\}$
so we need, given arbitrary continuous map $g:T\times\mathbb{Q}\to\mathbb{R}$ for $\{t\in T : g(\{t\}\times A)\subseteq U\}$ to be open
but tube lemma suggests this can't be
or does it
$g^{-1}(U)$ isn't just arbitrary open set, thats kind of the point
Is this proof correct: Question: Suppose that $G$ is a graph on $n$ vertices that contains no subgraph isomorphic to $F$. Suppose also that the number of independent sets of size $t$ in $G$ is at most $n^{t/2}$. Prove that the Ramsey number $r(F, K_t) \geq \sqrt{n}$.
Solution: Consider a random set $S$ by choosing each vertex with probability $p$ independently. Let $X = |S|$ and $Y$ be the number of occupied edges. For a fixed edge $e \in E$, define $Y_e$ as the indicator variable that all vertices in $e$ are chosen. The probability that $Y_e = 1$ is $p^t$.
By linearity of expectations, we have $\mathbb{E}(X) = np$ and $\mathbb{E}(Y) = \sum_{e \in E} \mathbb{E}(Y_e) = mp^t$, where $m$ is the total number of edges. The alteration involves removing one vertex from each edge, resulting in an independent set $I$. Thus, $\mathbb{E}(|I|) = \mathbb{E}(X) - \m…
