33
votes
Accepted
What is the smallest set of real continuous functions generating all rational numbers by iteration?
It is enough with one continuous function. First, I'll give a simple example with one function which is discontinuous at one point. To do it, consider the function $$f:(0,\pi+1)\to(0,\pi+1)$$ with
$$
...
- 10.6k
26
votes
Quantifier complexity of the definition of continuity of functions
It is truly a very nice question, one of those questions with an answer one feels must be right, but it is not so clear at first how to prove it.
Nevertheless, aiming at partial progress, I claim that ...
- 236k
22
votes
What is the smallest set of real continuous functions generating all rational numbers by iteration?
You only need one continuous function.
There exists a continuous function $f: \mathbb{R} \to \mathbb{R}$ with a dense orbit, according to this MathOverflow answer. As in Saúl's construction, you can ...
- 6,571
21
votes
Accepted
Function whose sets of discontinuities and zeros are the rationals
There isn't such a function. If $f$ is nonzero and continuous at some point $x$, then there is a neighbourhood of $x$ on which $f$ doesn't vanish. Hence if the set of zeros of $f$ is dense, then the ...
- 28.2k
19
votes
Accepted
Is $\mathbb{Q}$ the orbit of a rational function under iteration?
As was mentioned in the comments by pregunton, it is possible to do using two rational functions. I claim it is not possible using just one. As Fedor Petrov suggests in another comment, this is ...
- 28.2k
15
votes
Accepted
Does the class of Hausdorff spaces have a shared "Coordinate space"?
First of all, let me point out that $[0,1]$ is not a coordinate space for the class of completely regular spaces.
The definition of completely regular spaces says that for any closed $C \subset X$ ...
- 18.5k
14
votes
Accepted
Injective uniformly continuous function $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}$?
No. A uniformly continuous function takes $O(N)$ distinct values on an $N\times N$ grid.
- 109k
14
votes
Accepted
What is this equivalence relation on topological spaces: there are bijective continuous maps in both directions
This relation was introduced (I don't know if for the first time) in the 1984 paper Bijectively related spaces I: Manifolds
by P. H. Doyle and J. G. Hocking. As the title indicates, two spaces that ...
- 11.5k
11
votes
Accepted
Brouwer's Theorem in the free topos?
To summarize, the Lambek and Scott book actually says that functions on the reals in the free topos represent continuous functions. The nLab previously made the stronger claim that Brouwer's Theorem ...
11
votes
Accepted
Does this Osgood-like condition imply continuity?
You do not need such heavy high-tech as Korn's inequality or even Lebesgue measure theory for an elementary geometry homework. Let's say $F(0)=0$.
The first claim is that $F$ is bounded in some ...
- 61.9k
11
votes
Accepted
A subcontinuous function, which is not continuous
Let $(e_n)$ be the standard orthonormal basis of $\ell^2$: recall that, as a sequence, $(e_n)$ converges weakly to $0$. Now define a map $f\colon\mathbb{R} \to \ell^2$ by $f(\frac{1}{n})=e_n$ and $f(...
- 32.5k
11
votes
Accepted
Non-homeomorphic connected one-dimensional Hausdorff spaces that have continuous bijections between them in both sides
Here's such a construction, actually producing an infinite family of such spaces, actually planar and locally compact, pairwise in continuous bijection in both directions but pairwise non-homeomorphic....
- 63.9k
11
votes
Twice continuously differentiable implied by existence of limit
This is more of a long comment than answer. First, the analogous statement for the first derivative is already non-trivial, although not very difficult, see Aull, Charles E. "The first symmetric ...
- 8,992
10
votes
Accepted
For this continuous non differentiable function $f$ How to determine $\sup\{a\}$ s.t $\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h^\alpha}=0$ for all $x$?
The supremum you are looking for is $-\frac{\ln(3/4)}{\ln(4)}$. I leave my proof below, although as user479223 mentions, this is the same as showing Hölder continuity of the Weierstrass function, ...
- 10.6k
8
votes
Accepted
affine vs lipschitz
For instance, consider the convex subset $E:=\{x\in\ell_\infty: 0\le x_k\le 2^{-k} \text{ for all } k\ge0 \}$ of $\ell_\infty$. By dominated convergence, $E$ is compact and its relative topology ...
- 60.5k
8
votes
Accepted
A functional equation in two complex variables
$Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether ...
- 61.9k
8
votes
Quantifier complexity of definition of compactness
Often the way you prove that something isn't formalizable in first-order logic is (ironically enough) with a compactness proof. This is how you show, for instance, that there isn't a first-order ...
- 12.4k
8
votes
Accepted
Does the uniform boundedness principle holds for multilinear maps as well?
$\newcommand{\om}{\omega}$Let me answer your specific question.
The proof is similar to that of the uniform boundedness principle for linear functionals, but here using the identity
\begin{equation}
\...
- 128k
7
votes
Brouwer's Theorem in the free topos?
Dear All: you must be precise on what you mean by Brouwer's theorem. The free topos is closed under many rules, but unlike the realizability topos, it is seldom closed under the internal implicative ...
- 71
7
votes
Are point sets of the same order type connected by continuous (order type)-preserving motion?
The answer is indeed No. The most economic example up to now i think is mentioned Suvorov's. Independently examples was constructed by P. Mani, B. Jaggi, B. Sturmfels, N. White
"Uniform oriented ...
- 1,482
7
votes
Continuous functions and infinity
Yes, in fact,
$$\inf_{\delta>0}\ \liminf_{n\to\infty}f(n\delta) =\liminf_{x\to+\infty}f(x).$$
Assuming w.l.o.g. $\liminf_{x\to+\infty}f(x)<\alpha<+\infty$, the open set $A=\{f<\alpha\}$ ...
- 60.5k
7
votes
Is $\mathbb{Q}$ the orbit of a rational function under iteration?
A rational function is as a self-map of $\mathbb P^1$. With that understanding, as was noted earlier, it is possible to generate all of the points $\mathbb P^1(\mathbb Q)$ by starting with the point $...
- 47.4k
7
votes
Accepted
Can a power series of several variables be discontinuous on a compact set if it converges in every point of this set?
The series $f(x,y)=y+xy+x^2y+x^3y+\dots$ converges to $0$ when $y=0$, and converges to $y/(1-x)$ when $|x|<1$. This function is not continuous at $(x,y)=(1,0)$.
- 55.9k
7
votes
For this continuous non differentiable function $f$ How to determine $\sup\{a\}$ s.t $\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h^\alpha}=0$ for all $x$?
This is just local Holder continuity.
Let $f$ be $\gamma$ Holder at $x$. If $\alpha<\gamma$ then
$$\lim\limits_{h\to0}\frac{|f(x+h)-f(x)|}{h^\alpha}\leq\lim\limits_{h\to0} \frac{C h^\gamma}{h^\...
- 1,904
6
votes
Accepted
Continuous functions and infinity
Yes, this is true and well known. One of the references I know is the problem book of B. Makarov, M. Goluzina, A. Lodkin and A. Podkorytov (Selected problems in real analysis, Translations of ...
- 109k
6
votes
Continuity concepts for correspondences
Yes, these are the so-called Vietoris topologies. The upper Vietoris topology has a subbase consisting sets of the form $\{F\in 2^Y\mid F\subseteq O\}$ with $O$ open and the lower Vietoris topology ...
6
votes
Accepted
Functions with at most linear growth at infinity: is the constant itself continuous?
The answer is no to both your hopes: it can happen that neither $M_{f_n}\to M_f$ nor $\sup_n M_{f_n}<+\infty$ hold, although $M_f<\infty$.
As a counter-example take
$$
f_n(x)=\max(0,n(x-n)).
$$
(...
- 5,380
6
votes
Accepted
Is $\mathbb{Q}$ the orbit of a continuous function that is computable when restricted to $\mathbb{Q}$?
Yes. This answer is based on the answers to your previous question.
Start with a computable ergodic map $T$ (D. Thomine constructs an example here). For every basic open neighborhood $B_i$, the set $...
- 3,073
6
votes
On the continuity of a Set-Valued function (correspondence)
You don't, in general. In the special case,
$$f\left( x\right) =\left\{ y\in \mathbb{R}^{m}:
x ^{T}y\leq 0\right\} \text{,}$$
you get an obvious lack of lower hemicontinuity at $0$.
6
votes
Accepted
A continuous injection from $[0,1]$ to $\mathbb{R}^2$
If $x(t_0)\notin [x(0),x(1)]$ for some $t_0\in (0,1)$, then for a small positive $c$ we have $z(t_0)\notin [z(0),z(1)]$ where $z=x+cy$. Thus $z$ is not monotone, therefore not injective. But two ...
- 109k
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