## 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.

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## 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 $$x? Sorry, but I don’t even know how to start. Any ideas and impulses? Posted on Categories proxies ## There are no other clopen sets in$\mathbb{R}$except for$\mathbb{R}$and$\emptyset$Proof attempt: Let there be another clopen set $$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? Posted on Categories proxies ## 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. Posted on Categories proxies ## 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. Posted on Categories proxies ## 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. Posted on Categories proxies ## 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! Posted on Categories proxies ## 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? Posted on Categories proxies ## 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 $$K$$.

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