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$ .

Proof verification that the sequence $x_n = \frac{1}{n}$ converges to every point of $\mathbb{R}$ on the cofinite topology

Let $ a \in \mathbb{R}$ . Then any open set $ U$ in the cofinite topology containing $ a$ is of the form $ U = \mathbb{R} – \{\alpha_1, \alpha_2, \cdots, \alpha_p\}$ for some $ p \in \mathbb{N}$ (and of course $ \alpha_i \neq a \ \forall 1 \leq i \leq p$ ). Now, take $ N = c \left(\displaystyle{\frac{1}{\min \alpha_i}}\right) + 1$ , where $ c(x)$ stands for the ceiling of $ x$ (i.e, $ c(2.1) = 3$ ). Then it’s clear that $ x_n \in U \ \forall n \geq N$ and we’re done.

Is this alright? I actually think I could do away with all of this and just make the argument that the tail of $ x_n$ is eventually contained in $ U$ since $ \mathbb{R} – U$ has only finitely many elements and $ \{x_n\}_{n \in \mathbb{N}}$ is infinite, but I’m not sure if that’s fine too.

If $f: \mathbb{R} \to \mathbb{R}$ is a continuous surjection, must it be open?


If $ f: \mathbb{R} \to \mathbb{R}$ is a continuous surjection, must it be open?

I think not. I proved if $ f: \mathbb{R} \to \mathbb{R}$ is an open continuous surjection, then $ f$ is a homeomorphism. So, if the question is true, every continuous surjection must be a homeomorphism. But, I didn’t find a counterexample. Can someone help me?

Show $\lim_{a \to 0} a \cdot \mu (\{ x \in \mathbb{R} : |f(x)| > a \}) = 0$ for $f \in L^1 (\mathbb{R})$, $a>0$


Let $ f \in L^1 (\mathbb{R})$ , and $ a > 0$ . Show

$ $ \lim_{a \to 0} a \cdot \mu (E_a) = 0 $ $

where $ E_a = \{ x \in \mathbb{R} : |f(x)| > a \}$ .


Try

Since

$ $ \int_\mathbb{R} |f| d\mu = \int_{E_a} |f| d\mu + \int_{\mathbb{R}\setminus E_a } |f| d\mu < \infty $ $

we have $ \int_{E_a} |f| d\mu < \infty$ , $ \forall a$ .

But I’m stuck at how I can relate this to find $ \mu(E_a)$ .

Any help will be appreciated.

If $A \times B$ is Lebesgue measurable in $\mathbb{R}^2$ and $B$ is Lebesgue measurable in $\mathbb{R}$ then $A$ is so?

Though it seems simple but I am struggling to find a proof for it being a starter to measure theoty. I think this statement is true and I am trying to prove it. Here is my trial: Write $ A= A \times \{0\}= \cup_i(A_{i} \times B_{i})$ ,

$ m_1(A)=m_2(A \times \{0\})=\cup_i m_1(A_i)m_1(B_i)$

I don’t know what to do next! Is my approach correct?

where $ m_1,m_2$ are 1 and 2 dimensional measures in $ \mathbb{R},\mathbb{R^2}$ respectively.

Proving $\inf_{a \in \mathbb{R}} \mathbb{E}|\xi -a| = \mathbb{E}|\xi – m\xi|$

Let $ \xi$ be a random variable form $ (\Omega, \mathcal{F}, \mathbb{P})$ . Than let $ m\xi$ be a median of random variable $ \xi$ . I need to prove $ $ \inf_{a \in \mathbb{R}} \mathbb{E}|\xi -a| = \mathbb{E}|\xi – m\xi|.$ $

So I know that $ \mathbb{P}(\xi<m\xi) \leq \frac{1}{2} \leq \mathbb{P}(\xi \leq m\xi)$ .

I also know that $ $ \inf_{a\in \mathbb{R}} \mathbb{E}(\xi-a)^2=\mathbb{D}\xi$ $ when $ $ \mathbb{D}\xi=\mathbb{E}(\xi-\mathbb{E}\xi)^2.$ $

Can I somehow use it here or how should I prove it?

Show that $f_n \rightarrow 0$ in $C([0, 1], \mathbb{R})$

I was given the following problem and was wondering if I was on the right track.

Let $ f_n(x) = \frac{1}{n} \frac{nx}{1 + nx}, \: 0 \le x \le 1$

Show that $ f_n \rightarrow 0$ in $ C([0, 1], \mathbb{R})$ .

I have this theorem that I figured I could use:

$ f_k \rightarrow f$ uniformly on A $ \iff$ $ f_k \rightarrow f$ in $ C_b$ .

In this case, $ C_b$ is the collection of all continuous functions on $ [0,1]$ . So if I can prove the function is uniformly continuous, this would prove that $ f_n \rightarrow 0$ . Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?

Thanks