Let $f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function. $\int_{a}^{\infty} f$ converges.Prove that $\lim_{x\to\infty} f(x)=0$


Let $ f:[a,\infty)\rightarrow \mathbb{R}$ be a uniformly continuous function in that range. $ \int_{a}^{\infty} f$ converges. Prove that $ \lim_{x\to\infty} f(x)=0$

Hint: Use the sequence $ F_n(x)=n\int_{x}^{x+\frac{1}{n}} f$ .

Honestly I have been trying to solve this one for some time but the hint really confuses me.

I have tried to mess around with $ F_n(x)$ a bit, for example by using the fundamental theorem but it still seems like such a random choice and I can’t make anything out of it.

Any guidance/explanations will be appreciated.

Please use the hint in the question.

There are no other clopen sets in $\mathbb{R}$ except for $\mathbb{R}$ and $\emptyset$

Proof attempt:

Let there be another clopen set $ S$ in which is a proper subset of $ \mathbb{R}$ . Hence, $ S^c \neq \emptyset $ .

We can assert the following statements:

  1. No point of $ S$ lies in $ S^c$ .
  2. No point of $ \overline{S}$ lies in $ S^c$ [Since the set $ S$ is closed, no point of the derived set of $ S$ is a member of $ S^c$ ].

  3. No point of $ S^c$ lies in $ S$ .

  4. No point of $ \overline{S^c}$ lies in $ S$ . [$ S$ is both open and closed. Hence its compliment, i.e. $ S^c$ is closed and open. (compliment of a closed set is an open set and vice versa).]

Therefore, $ S \cap \overline{S^c} =\emptyset$ and $ \overline{S} \cap S^c= \emptyset $ . Therefore, $ S$ and $ S^c$ are separated sets. Again, $ S\cup S^c =\mathbb{R}$ . Therefore, $ \mathbb{R}$ is disconnected, a contradiction.

[ $ \mathbb{R}$ is connected, since for any $ x, y \in \mathbb{R}$ $ \implies$ $ z \in \mathbb{R}$ , where $ z$ is any point such that $ x<z<y$ .]

Is it correct?

What is a topology on $\mathbb{R}$ with a disconnected subset $A$ where the union of seperations of $A$ with $\mathbb{R}-A$ is not $\varnothing$?

$ \newcommand{\R}{\mathbb{R}}$

Hello, I was wondering how to find a topology on $ \R$ and a disconnected subset $ A$ such that every pair of sets, $ U$ and $ V$ , that is a separation of $ A$ in $ \R$ satisfies $ U \cap V \cap (\R – A ) \ne \varnothing$ ?

I was thinking the lower limit topology on $ \R$ would be such a topology.

For example, if $ \R = (-\infty, \infty), A = (-\infty,0), U = (-\infty, 0)\cup[1,\infty), V = [0,\infty)$ .

Then

  • $ A\subset U\cup V$
  • $ A\cap U\ne\varnothing$
  • $ A\cap V\ne\varnothing$
  • $ U\cap V\cap A=\varnothing$

So $ U$ and $ V$ are indeed a seperation of $ A$ and $ A$ is disconnected in $ \R$ .

And also we have $ U\cap V\cap(\R-A)\ne\varnothing$ , which is what we wanted.

But I am unsure if that works in all cases. We need to show this holds for all $ U$ and $ V$ pairs that are a separation of $ A$ .

Any help would be much appreciated! Thanks.

What is the cardinal of the set of Uniform Continous functions with domain in $\mathbb{R}$?

I learned in my topology class that $ \mathbb{R} \cong \mathfrak{C}(\mathbb{R})$ The latter being the set of continous functions with domain in $ \mathbb{R}$ .

Since if a function is uniformly continous then it is continous, the set of all uniform function must have cardinal $ \leq \mathfrak{c}$ .

But continous functions need not be uniformly continous; does this show that there is no bijection from continous functions to uniformly continous functions?

Im kind of lost with all this cardinality stuff, its not intuitive at all.

Bijection $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$

I need a bijection, such that: $ f(x,y) = \phi(x) + \psi(y)$ and $ f(x,y) = g(\phi(x) + \psi(y))$ . Second one is easy, I think. If we make first one, we can consider $ g$ as identical map. I have seen similar topics on stack, but I don’t like those answers. I think it is possible to construct a bijection, as Cantor did, proving that $ [0,1] \times [0,1]$ ~ $ [0,1]$ , because I can build a bijection $ h: [0,1] \to \mathbb{R}$ . Thus we have a equivalent problem, with segment, i.e build a bijection $ f: [0,1] \times [0,1] \to [0,1]$ , such that $ f(x,y) = \phi(x) + \psi(y)$ .

Thank you in advance!

What is the fastest algorithm to establish whether a linear system in $\mathbb{R}$ has a solution?

I know the best algorithm to solve a linear system in $ \mathbb{R}$ with $ n$ variables is Coppersmith-Winograd’s algorithm, which has a complexity of $ $ O\left(n^{2.376}\right). $ $ How much easier is it to simply determine whether the same system has any solution?

More precisely, given a system of $ m$ equations and $ n$ unknowns, what is the complexity of establishing whether it has any solution?

Example of a polynomial that fails to be non-zero over a higher field than $\mathbb{C}$ or $\mathbb{R}$?

It’s well known over the field $ \mathbb{F}_2 $ that the polynomial $ p(x) = x^2 – x$ is equivalent to the $ 0$ polynomial over the field. Naturally however if we consider say $ \mathbb{R}$ then it’s easy to see that $ x^2 -x$ is not $ 0$ everywhere.

This leads to an interesting question: are there examples of (possibly multivariate) polynomial expressions $ P(x_0,x_1…x_k)$ such that $ P$ is $ 0$ everywhere over $ \mathbb{C}$ or $ \mathbb{R}$ but such that there is a field $ K$ of which $ \mathbb{C}$ or $ \mathbb{R}$ is a quotient field (or sub-field), and the specific polynomial $ P$ is not identically $ 0$ over the entirety of $ K$ .