@DiscipleofBarney for step 5 their is something I don't get here we at first we constructed a number $\mu_{U}(K)$ that depends on the open set U now we do it to make it construct a number on u(K) without requiring specific open set of U however it says at the end we may pick some u $\in$ intersection of C(V)
@DiscipleofBarney understand when he says in step 6 thinking of elements of X as functions from K to R
I mean for any arbitrarily cartesian product we can think of it as a function from
X to the union
however I don't see why this is a composition of function
I understand that we can think of cartesian product as a function f : X $\rightarrow$ U($X_i$) such that f($X_i$) $\in$ $X_i$ , so here we will have the union will be elements of the real numbers.
I don't understand that step at all.
I mean we showed that $\mu_{U}($K_1$)$ $\leq$ $\mu_U(K_2)$ however why do we even need to do this since this happens for every element u in particular we should have $\mu$($K_1$) $\leq$ $\mu$($K_2$)
?
nvm I understood it
I still don't understand why its composition of maps
@KarimMansour First, these questions seem like you basically want me to explain the whole proof which I don't really feel like going through in its entirety. I am fine with the questions, just be warned that I am just going by the "local" context so an explanation I give you might actually not be what the author is referring to.
It looks like the composition of maps things is saying that since $f$ is continuous, then $f \times f$ is continuous, and minus is continuous so the function $X \to \mathbb{R}$ is the composition of $f \mapsto f \times f \mapsto f - f$
Well it is proven that is the case, but $ \mu$ is in the closures, so it might not be of the form $\mu_U$ so you need continuity to also show that the inequality extends to $\mu$.
Thats like having a function that is continuous on the rationals, but I can extend the function in a bunch of non continuous ways on the reals, so we need some restriction on how everything interacts
@DiscipleofBarney still don't entirely get the elements of function of X I understand now why we need to see if the map f($K_2$) - f($K_2$) is continuos. But I don't understand the composition itself.
@KarimMansour Its actually $f \mapsto (f,f)$ now that I think about it, so $$f \mapsto (f,f) \mapsto f - f \mapsto f(K_1) -f(K_2)$$, its just a fancy way of saying that if $f,g$ are continuous real valued functions then $f-g$ is also continuous.
It comes up in topology, sometimes analysis, basically giving standard constructions on how to construct continuous functions from other continuous functions
" It follows that the restricts to a measure on the Borel subsets of G , so that it is a Borel measure ( G is completely regular, as mentioned early, and in particular Hausdorff)"
I'm interested to see how this turns out. My intuition tells me we'll be integrating $g(x) = x$, just "shuffled", but I'm probably wrong about that. Possibly just $n\int_0^1 x dx$
So to really maximize on Fridays/Saturdays we do the 40 extra vote thing, then vote on as many closed questions, then vote again. If you tried really hard you could probably get like 100votes on those days
Yah, Mabe someone needs to come up with a stackexchange site where they votes up all the bad questions, so that they can be easily found on the real site @Incurrence
I have wondered about that too, maybe accidentlaly upvoted and it timed out so couldn't take the votes back, must happen every once in a while with that many downvotes
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Fundamental I have from the 6th to the 22nd ish from memory
@robjohn Yesterday I discovered a new double integral representation of Catalan's constant. I checked more sources and no one says anything about that representation. I think I'll add it to my book, no publishing before.
Background
I am trying to solve the following problem:
>
Given 2 distinct curves $C_1: y=f(x)=e^{6x}$ and $C_2: y=g(x)=ax^2$ where $a>0$. The objective is to find the range of $a$ such that there exists 2 tangents, each is tangent to both given curves.
I have solved it as follows.
Let ...
@Incurrence there is a general expression for arcsin(z) as log(something). try fiddling with the Euler's formula for a while, you'll be able to derive it.
@Balarka tell me something if all the elements in set B are mapped onto by some elements in set A leaving some elements in set A this is a bijective function right?
What do 'branches' refer? Is this referring to, for example with a hyperbola, we have two branches, left and right(if x^2-y^2) and top and bottom branches with (y^2-x^2)?
i took a class on alg top in the uni i often go to. only recently gave up as they started cohomology. thought i'd rather learn it in a De Rham-ish way.
Background
I am trying to solve the following problem:
>
Given 2 distinct curves $C_1: y=f(x)=e^{6x}$ and $C_2: y=g(x)=ax^2$ where $a>0$. The objective is to find the range of $a$ such that there exists 2 tangents, each is tangent to both given curves.
I have solved it as follows.
Let ...
