## 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.

## How to prove $(\mathbb{R}\backslash \mathbb{Q})\cap (x,y)\neq \emptyset$ for $x,y\in \mathbb{R}$ and $x How to prove $$(\mathbb{R}\backslash \mathbb{Q})\cap (x,y)\neq \emptyset$$ for $$x,y\in \mathbb{R}$$ and $$x? Sorry, but I don’t even know how to start. Any ideas and impulses? ## There are no other clopen sets in$\mathbb{R}$except for$\mathbb{R}$and$\emptyset$Proof attempt: Let there be another clopen set $$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.] Is it correct? ## uniform convergence of series on$\mathbb{R}$For the series $$\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})$$, I can prove that it converges uniformly on every compact subset of $$\mathbb{R}$$. Does it converges uniformly on $$\mathbb{R}$$ though? I don’t think so, so I’m trying to find a $$x$$ that will cause the series to diverge but I can’t find any. ## Interesting subsets of$\mathbb{R}$When thinking about subsets of the reals $$\mathbb{R}$$, intervals immediatly come to mind. What other interesting subsets are there which can be used as counterexamples (e.g. the Cantor set) or are handy to construct certain examples? ## 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 $$K$$.