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10 votes
5 answers
2k views

Extracting a common convergent indexing from an uncountable family of sequences

Let $\mathcal{A}$ be some uncountable index set and $X$ be some separable reflexive Banach space. For each $\alpha \in \mathcal{A}$, let \begin{equation} \{ x_n^{\alpha} \}_{n=1}^\infty \end{equation} ...
Isaac's user avatar
  • 3,477
3 votes
1 answer
394 views

Approximations for real functions

Is there some set of real functions ($S$) that has precisely the cardinality of the continuum, but is dense in the space of all the real functions in a sense that every real function can be ...
TeeZee's user avatar
  • 39
3 votes
2 answers
2k views

Can every real function be approximated with a Riemann-integrable one with any precision required?

Is there some proof that Riemann-integrable functions are dense in the space of all real functions? In a sense that for every real function $f$ and number $\varepsilon>0$, there is Riemann-...
user479568's user avatar
0 votes
0 answers
94 views

Is the space of affine continuous functions a Baire space

Let $\Omega$ be a compact convex set in q linear normed space. Let $A(\Omega)$ be the space of affine continuous real-valued functions. My question is whether the space $A(\Omega)$ is a Baire space? ...
user119197's user avatar
9 votes
1 answer
831 views

Baire category theorem for uncountable unions

Any compact Hausdorff space $X$ is a Baire space: if the set $X$ is a meager set (meaning a countable union of nowhere dense subsets, also known as a set of first category), then $X$ is empty. I am ...
Dmitri Pavlov's user avatar
4 votes
1 answer
224 views

Bounded growth of functions vs bounded growth of functions on countable sets

I am wondering if the boundedness of growth can be characterized by sequences. I am not sure if I use the term "growth" correctly, or use the correct tags for this question. Here is what I mean. Let $...
erz's user avatar
  • 5,529
7 votes
2 answers
665 views

Non-separable metric probability space

Let us say a metric probability space $(X,\rho,\mu)$ has property (*) if: the support of $\mu$ is contained in a separable subspace of $X$. Questions: 1. Is there a standard name for this property? ...
Aryeh Kontorovich's user avatar
5 votes
1 answer
795 views

How to define transfinite derivatives of a function?

There are all manners of theories generalizing the notion of derivative. Amongst them is the fractional calculus, a rich theory which gives a sense to the derivation and integration of non-integer (i....
Morteza Azad's user avatar
11 votes
1 answer
766 views

Generalized limits on $\ell^\infty(\mathbb{N})$

Let $\ell^\infty(\mathbb{N})$ denote the set of bounded real sequences $(a_n)_{n\in\mathbb{N}}$. The $\lim$ operator is a partial linear operator from $\ell^\infty(\mathbb{N})$ to $\mathbb{R}$. With ...
Dominic van der Zypen's user avatar
1 vote
1 answer
242 views

Can (how) one distinguish germs of continuous functions by a countable set of params?

Continuous functions can be distinguished by their values at say rational points of [0 1]. Germs of analytic functions can be distinguished by derivatives at a point. So in both cases we see ...
Alexander Chervov's user avatar
7 votes
3 answers
4k views

Is a semicontinuous real function Borel measurable?

Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function. [Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example? Note that, for any $c$, ...
kenneth's user avatar
  • 1,399
26 votes
2 answers
5k views

Does Arzelà-Ascoli require choice?

Inspired by a recent Math.SE question entitled Where do we need the axiom of choice in Riemannian geometry?, I was thinking of the Arzelà--Ascoli theorem. Let's state a very simple version: ...
Nate Eldredge's user avatar