Dtjytyk

How should I respond to a player wanting to catch a sword between their hands?

If A makes B more likely then B makes A more likely"

3 doors, three guards, one stone

Fishing simulator

Was credit for the black hole image misattributed?

Who can trigger ship-wide alerts in Star Trek?

Two different pronunciation of "понял"

What are the performance impacts of 'functional' Rust?

What was the last x86 CPU that did not have the x87 floating-point unit built in?

What did Darwin mean by 'squib' here?

Is it possible to ask for a hotel room without minibar/extra services?

Area of a 2D convex hull

New Order #5: where Fibonacci and Beatty meet at Wythoff

Why does this iterative way of solving of equation work?

Notation for two qubit composite product state

Why is "Captain Marvel" translated as male in Portugal?

Why does tar appear to skip file contents when output file is /dev/null?

How to rotate it perfectly?

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

What's the difference between (size_t)-1 and ~0?

Need a suitable toxic chemical for a murder plot in my novel

Windows 10: How to Lock (not sleep) laptop on lid close?

How to colour the US map with Yellow, Green, Red and Blue to minimize the number of states with the colour of Green

Antler Helmet: Can it work?

Is this graph dense?

0

$begingroup$

I am trying to find a function $[0,1]to[0,1]$ whose graph is dense in $[0,1]times[0,1]$ and came up with this:

Let $f:[0,1]longrightarrowmathbb{R}$ be defined as follows: if $x$ is irrational then $f(x)=0$ and if $x=p/q$ is an irreducible rational then $f(x)=q$. Now $F(x)=sin(f(x)$) is the required function (?). This are some questions that come to mind:

Let $f:mathbb{R}tomathbb{R}$ be a bounded function such that ${f(n): ninmathbb{N}}subseteq[a,b]$ is dense (in $[a,b]$) and let $g:[0,1]tomathbb{R}$ be another bounded function such that $g(1/n):= f(n)$. Is the segment ${0}times[a,b]$ in the closure of $Graph(g)={(x,g(x)):xin(0,1]}$? I think this should be true and not so hard to prove but does the result depend on the distribution of $f(n)$ in $[a,b]$? For example, $ sin(n)$ has density $frac{1}{pisqrt{1-x^2}}$. Does the result hold only when the distribution is uniform or whenever the density is non-zero? I'm inclined for the latter.

The idea in $sin(f(x))$ is that $f$ blows up to infinity in every irrational and it reaches every single natural number greater than some constant $K$ (depending on the irrational) so ${sin(f(x)):xtext{ close to some irrational }}={sin(n):ninmathbb{N};nge K}$ is still dense, which would imply that $mathbb{I}times[-1,1]$ (and its closure) are contained in the closure of $Graph(F)$, but its closure is $[0,1]times[-1,1]$ because the irrationals are dense in $[0,1]$. It is like making "${0}times[-1,1]$" is in the closure of $Graph(sin(1/x))$" but for a lot of points (a dense set of them). Does it make sense?

general-topology graphing-functions

share|cite|improve this question

asked Mar 23 at 11:42

PedroPedro

541212

$endgroup$

add a comment | 

0

$begingroup$

I am trying to find a function $[0,1]to[0,1]$ whose graph is dense in $[0,1]times[0,1]$ and came up with this:

Let $f:[0,1]longrightarrowmathbb{R}$ be defined as follows: if $x$ is irrational then $f(x)=0$ and if $x=p/q$ is an irreducible rational then $f(x)=q$. Now $F(x)=sin(f(x)$) is the required function (?). This are some questions that come to mind:

Let $f:mathbb{R}tomathbb{R}$ be a bounded function such that ${f(n): ninmathbb{N}}subseteq[a,b]$ is dense (in $[a,b]$) and let $g:[0,1]tomathbb{R}$ be another bounded function such that $g(1/n):= f(n)$. Is the segment ${0}times[a,b]$ in the closure of $Graph(g)={(x,g(x)):xin(0,1]}$? I think this should be true and not so hard to prove but does the result depend on the distribution of $f(n)$ in $[a,b]$? For example, $ sin(n)$ has density $frac{1}{pisqrt{1-x^2}}$. Does the result hold only when the distribution is uniform or whenever the density is non-zero? I'm inclined for the latter.

The idea in $sin(f(x))$ is that $f$ blows up to infinity in every irrational and it reaches every single natural number greater than some constant $K$ (depending on the irrational) so ${sin(f(x)):xtext{ close to some irrational }}={sin(n):ninmathbb{N};nge K}$ is still dense, which would imply that $mathbb{I}times[-1,1]$ (and its closure) are contained in the closure of $Graph(F)$, but its closure is $[0,1]times[-1,1]$ because the irrationals are dense in $[0,1]$. It is like making "${0}times[-1,1]$" is in the closure of $Graph(sin(1/x))$" but for a lot of points (a dense set of them). Does it make sense?

general-topology graphing-functions

share|cite|improve this question

asked Mar 23 at 11:42

PedroPedro

541212

$endgroup$

add a comment | 

$begingroup$

I am trying to find a function $[0,1]to[0,1]$ whose graph is dense in $[0,1]times[0,1]$ and came up with this:

Let $f:[0,1]longrightarrowmathbb{R}$ be defined as follows: if $x$ is irrational then $f(x)=0$ and if $x=p/q$ is an irreducible rational then $f(x)=q$. Now $F(x)=sin(f(x)$) is the required function (?). This are some questions that come to mind:

Let $f:mathbb{R}tomathbb{R}$ be a bounded function such that ${f(n): ninmathbb{N}}subseteq[a,b]$ is dense (in $[a,b]$) and let $g:[0,1]tomathbb{R}$ be another bounded function such that $g(1/n):= f(n)$. Is the segment ${0}times[a,b]$ in the closure of $Graph(g)={(x,g(x)):xin(0,1]}$? I think this should be true and not so hard to prove but does the result depend on the distribution of $f(n)$ in $[a,b]$? For example, $ sin(n)$ has density $frac{1}{pisqrt{1-x^2}}$. Does the result hold only when the distribution is uniform or whenever the density is non-zero? I'm inclined for the latter.

The idea in $sin(f(x))$ is that $f$ blows up to infinity in every irrational and it reaches every single natural number greater than some constant $K$ (depending on the irrational) so ${sin(f(x)):xtext{ close to some irrational }}={sin(n):ninmathbb{N};nge K}$ is still dense, which would imply that $mathbb{I}times[-1,1]$ (and its closure) are contained in the closure of $Graph(F)$, but its closure is $[0,1]times[-1,1]$ because the irrationals are dense in $[0,1]$. It is like making "${0}times[-1,1]$" is in the closure of $Graph(sin(1/x))$" but for a lot of points (a dense set of them). Does it make sense?

general-topology graphing-functions

share|cite|improve this question

asked Mar 23 at 11:42

PedroPedro

541212

$endgroup$

share|cite|improve this question

asked Mar 23 at 11:42

PedroPedro

541212

share|cite|improve this question

asked Mar 23 at 11:42

PedroPedro

541212

asked Mar 23 at 11:42

PedroPedro

541212

asked Mar 23 at 11:42

PedroPedro

541212

1 Answer
1

active

oldest

votes

1

$begingroup$

To find functions with dense graphs, go more extreme, e.g.'s Conway's base 13 function, or non-constructive ones that can be found by transfinite recursion and that have the property that $f[(a,b)]=mathbb{R}$ for every $a < b$. (take the sine of it, if the image of $[0,1]$ is important)

Your proposed function $F$ is not good enough I think.

share|cite|improve this answer

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159218%2fis-this-graph-dense%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

To find functions with dense graphs, go more extreme, e.g.'s Conway's base 13 function, or non-constructive ones that can be found by transfinite recursion and that have the property that $f[(a,b)]=mathbb{R}$ for every $a < b$. (take the sine of it, if the image of $[0,1]$ is important)

Your proposed function $F$ is not good enough I think.

share|cite|improve this answer

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

1

$begingroup$

To find functions with dense graphs, go more extreme, e.g.'s Conway's base 13 function, or non-constructive ones that can be found by transfinite recursion and that have the property that $f[(a,b)]=mathbb{R}$ for every $a < b$. (take the sine of it, if the image of $[0,1]$ is important)

Your proposed function $F$ is not good enough I think.

share|cite|improve this answer

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

add a comment | 

$begingroup$

To find functions with dense graphs, go more extreme, e.g.'s Conway's base 13 function, or non-constructive ones that can be found by transfinite recursion and that have the property that $f[(a,b)]=mathbb{R}$ for every $a < b$. (take the sine of it, if the image of $[0,1]$ is important)

Your proposed function $F$ is not good enough I think.

share|cite|improve this answer

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

$endgroup$

share|cite|improve this answer

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127

answered Mar 23 at 16:06

Henno BrandsmaHenno Brandsma

116k349127