33 votes
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can another topology be given to $\mathbb R$ so it has the same continuous maps $\mathbb R\rightarrow \mathbb R$?

The only topology similar to the Euclidean topology on $\mathbb{R}$ is the Euclidean topology. Suppose there is such a topology $\tau$. I'll use "open," "continuous," etc. to mean with respect to ...
Gabriel C. Drummond-Cole's user avatar
33 votes
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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 $$ ...
Saúl RM's user avatar
  • 7,916
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 ...
Joel David Hamkins's user avatar
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 ...
Martin M. W.'s user avatar
  • 5,536
21 votes
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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 ...
Wojowu's user avatar
  • 27.4k
19 votes
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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 ...
Wojowu's user avatar
  • 27.4k
15 votes

can another topology be given to $\mathbb R$ so it has the same continuous maps $\mathbb R\rightarrow \mathbb R$?

This is a special case of much more general results surveyed in the book MR0393330 Magill, K. D., Jr. A survey of semigroups of continuous selfmaps. Semigroup Forum 11 (1975/76), no. 3, 189–282. For ...
Alexandre Eremenko's user avatar
15 votes
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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 $C \subset X$ ...
Will Brian's user avatar
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14 votes
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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.
Fedor Petrov's user avatar
14 votes
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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 ...
Ramiro de la Vega's user avatar
11 votes
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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 ...
fedja's user avatar
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11 votes
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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(...
Gro-Tsen's user avatar
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11 votes
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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....
YCor's user avatar
  • 60.1k
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 ...
Kostya_I's user avatar
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10 votes
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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 ...
9 votes
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Continuous non-constant function with infinite intersections with horizontal line on a compact interval?

Preimage $f^{-1}(v)$ of any value $v$ is a closed set, hence its complement $U(v)$ is open. This open set $U(v)$ is a disjoint union of intervals. If some interval is finite, say $(a,b)$, then $f(a)=f(...
Fedor Petrov's user avatar
8 votes
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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 ...
Pietro Majer's user avatar
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8 votes
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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 ...
fedja's user avatar
  • 59.5k
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 ...
James Hanson's user avatar
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8 votes
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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} \...
Iosif Pinelis's user avatar
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 ...
Philip Scott's user avatar
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 ...
Nikolai Mnev's user avatar
  • 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 $A=\{f<\alpha\}$ ...
Pietro Majer's user avatar
  • 56.5k
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 $...
Joe Silverman's user avatar
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 $(x,y)=(1,0)$.
Tom Goodwillie's user avatar
6 votes
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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 ...
Fedor Petrov's user avatar
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 ...
Michael Greinecker's user avatar
6 votes
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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 $$ f_n(x)=\max(0,n(x-n)). $$ (...
leo monsaingeon's user avatar
6 votes
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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 $...
Dan Turetsky's user avatar
  • 2,678
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 $0$.
Michael Greinecker's user avatar

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