If you multiply your function by a linear transformation the derivate is also multiplied by that linear transformation. So if you multiply it by the inverse of its derivative, you get a new function g whose derivative is the identity map.
Here's a sketch I would do, lots of details left to you
1.) if ##t_0\in A## then ##t\in A \forall t < t_0##.
2.) if ##t_k\to t## is a sequence with ##t_k\in A## for all k, then ##t\in A## unless ##t_k< t## for all k by (1)
3.) you can have pick a strictly increasing subsequence (unless ##t=t_k##...
Proving the set of indices not in the subsequence converges is really roundabout. Given ##\epsilon>0## there exists ##n_k## such that ##|1/2-f_{n_k}|<\epsilon##. What can you say about ##f_n## for any ##n>n_k##? Don't think about whether it's part of the subsequence or not.
It's an annoying quirk that you need double # instead of single $ to make in line latex, if you could edit your post to find and replace all instances it would be easier to read
I think you only need one turn? Assuming your arms can have arbitrary length, the first arm can reach the (x,y,0) point corresponding to any (x,y,z), then you just make a single turn up or down and attach another arm of length z.
Are the lengths of the arms restricted in any way? E.g. the set...
A good starting point is to compare it to a geometric series. For example if ##c=1/3## can you think of a series that converges whose terms are eventually guaranteed to be larger than the ##x_n##?
Suppose ##f## is continuous and ##f(A)=B##, where ##A## is compact. Let ##U_i## be an open infinite cover of ##B## and consider ##V_i=f^{-1}(U_i)##. This is an open subcover of ##A## so has a finite subcover which I will call ##V'_i##. Let ##U'_i## be the subset of ##U_i## for which...
I'm going to do things over two dimensions, to make the difference between scalars and vectors a bit more obvious. ##\mathbb{Q}^2## is obviously a 2 dimensional vector space over ##\mathbb{Q}##. It is also a subset of ##\mathbb{R}^2##. This is both a two dimensional vector space over...
No, for example the sequence ##x_k=(1,0)## when ##k## is odd and ##(0,1)## when ##k## is even does not converge.
But it is true that every sequence contained in a closed bounded set has a convergent subsequence. And that subsequence will give you your contradiction.
Item 1 does not exist.
The key point is for the supremum to be 1 but the maximum to be less than 1, there must exist a sequence ##x_n## such that ##f(x_n)\to 1## (probably worth proving if it's not obvious) What do you know about sequences in the closed unit balls? Are they guaranteed to have...
Counterpoint: ##\mathbb{Q}\subset \mathbb{R}##, is a 1 dimensional vector space over itself but it's not a subspace of the one dimensional real vector space ##\mathbb{R}##. I would argue the addition is the same :)
You can kind of just compute this exactly. For any possible choice of x, you know what fraction of the time your sampling will return a 0 instead of a 1. Then you can compute things like what value of x makes it so you would only see at least as extreme a result as you got 5% of the time (in...