Xjyuk

What is the delta-v required to get a mass in Earth orbit into the sun using a SINGLE transfer?

Did the Byzantines ever attempt to move their capital to Rome?

Automatically end list item with proper punctuation (semicolon, period)

Dragons and gems

Complex conjugate and transpose "with respect to a basis"

Why can't some airports handle heavy aircraft while others do it easily (same runway length)?

I multiply the source, you (probably) multiply the output!

Template default argument loses its reference type

What precisely does the commonly reported network hash rate refer to?

The Green Glass Door, Revisited

Word for something that used to be popular but not anymore

Can taking my 1-week-old on a 6-7 hours journey in the car lead to medical complications?

Could someone please explain what this inline #define assembly is doing?

Why is infinite intersection "towards infinity" an empty set?

Why do the Brexit opposition parties not want a new election?

Is there a "right" way to interpret a novel, if not, how do we make sure our novel is interpreted correctly?

Why did Tony's Arc Reactor do this?

Why does low tire pressure decrease fuel economy?

is it possible to change a material depending on whether it is intersecting with another object?

Would a character take damage when surrounded by, but not in, flames?

Was Robin Hood's point of view ethically sound?

How to finish my PhD?

Capacitors with same voltage, same capacitance, same temp, different diameter?

If every star in the universe except the Sun were destroyed, would we die?

Why is infinite intersection “towards infinity” an empty set?

Questions regarding the Proof of Egorov's Theorem (Carothers)Stable set by intersection and by finite unionExample of decreasing sequence of sets with first set having infinite measureEventually in a set, meaning?Infinite Union/Intersection vs Infinite summationinfinite set don't contain sigma algebra that have more than one countable complementCan the interior of the intersection of a closed and an open set be empty if the intersection is non-empty?Why is an uncountable union of null sets not necessarily a null set?Complex measure of empty set.

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

$begingroup$

Why is infinite intersection "towards infinity" an empty set?

Or i.e.

Why is:

$$cap_i=1^infty F_n = emptyset$$

$$F_n=[n, infty)$$

There's intuition, the intersection is always the "smallest of the sets" so eventually it will be $(infty,infty)$ or something like that.

measure-theory

share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Why is infinite intersection "towards infinity" an empty set?

Or i.e.

Why is:

$$cap_i=1^infty F_n = emptyset$$

$$F_n=[n, infty)$$

There's intuition, the intersection is always the "smallest of the sets" so eventually it will be $(infty,infty)$ or something like that.

measure-theory

share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

$endgroup$

add a comment | 

$begingroup$

Why is infinite intersection "towards infinity" an empty set?

Or i.e.

Why is:

$$cap_i=1^infty F_n = emptyset$$

$$F_n=[n, infty)$$

There's intuition, the intersection is always the "smallest of the sets" so eventually it will be $(infty,infty)$ or something like that.

measure-theory

share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

asked 8 hours ago

mavaviljmavavilj

2,96311 gold badge1313 silver badges4343 bronze badges

4 Answers
4

active

oldest

votes

6

$begingroup$

The intersection is made up of real numbers which are greater than or equal to every positive integer. By Archimedes' property, there's none.

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

To belong in the intersection, any element would have to belong in each of the sets $[n,infty)$ which means it must be larger than every finite number. Since there is no finite number with this property, the intersection is therefore empty.

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Suppose, to obtain a contradiction, that $F=cap_n F_n$ is non-empty. Let $x in F$. Then, $xge n$ for all $n in mathbb N$. However, there is no largest real number, so we must conclude that $F = emptyset$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

It might be a little easier to understand via the contrapositive:

Let $x$ be any real number. Then we know that there exists a positive integer $n_x$ larger than $x$. (This is the so-called Archimedean property of the reals, but intuitively you can think of $n_x$ just being $x$ rounded up to the next integer, or if $x$ is negative you can just let $n_x$ be $1$.) That means $x$ is not in the set $F_n_x$, so it certainly cannot be in the intersection $bigcap_n F_n$. This is true no matter what $x$ is, so $bigcap_n F_n$ cannot contain any real numbers at all, which is to say it equals the empty set.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

$endgroup$

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/4.0/"u003ecc by-sa 4.0 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%2f3347553%2fwhy-is-infinite-intersection-towards-infinity-an-empty-set%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

The intersection is made up of real numbers which are greater than or equal to every positive integer. By Archimedes' property, there's none.

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

The intersection is made up of real numbers which are greater than or equal to every positive integer. By Archimedes' property, there's none.

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

$endgroup$

add a comment | 

$begingroup$

The intersection is made up of real numbers which are greater than or equal to every positive integer. By Archimedes' property, there's none.

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

3

$begingroup$

To belong in the intersection, any element would have to belong in each of the sets $[n,infty)$ which means it must be larger than every finite number. Since there is no finite number with this property, the intersection is therefore empty.

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

To belong in the intersection, any element would have to belong in each of the sets $[n,infty)$ which means it must be larger than every finite number. Since there is no finite number with this property, the intersection is therefore empty.

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

$endgroup$

add a comment | 

$begingroup$

To belong in the intersection, any element would have to belong in each of the sets $[n,infty)$ which means it must be larger than every finite number. Since there is no finite number with this property, the intersection is therefore empty.

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

answered 8 hours ago

pre-kidneypre-kidney

18.3k2222 silver badges5959 bronze badges

1

$begingroup$

Suppose, to obtain a contradiction, that $F=cap_n F_n$ is non-empty. Let $x in F$. Then, $xge n$ for all $n in mathbb N$. However, there is no largest real number, so we must conclude that $F = emptyset$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Suppose, to obtain a contradiction, that $F=cap_n F_n$ is non-empty. Let $x in F$. Then, $xge n$ for all $n in mathbb N$. However, there is no largest real number, so we must conclude that $F = emptyset$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

$endgroup$

add a comment | 

$begingroup$

Suppose, to obtain a contradiction, that $F=cap_n F_n$ is non-empty. Let $x in F$. Then, $xge n$ for all $n in mathbb N$. However, there is no largest real number, so we must conclude that $F = emptyset$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

answered 8 hours ago

Theoretical EconomistTheoretical Economist

3,98322 gold badges88 silver badges3131 bronze badges

1

$begingroup$

It might be a little easier to understand via the contrapositive:

Let $x$ be any real number. Then we know that there exists a positive integer $n_x$ larger than $x$. (This is the so-called Archimedean property of the reals, but intuitively you can think of $n_x$ just being $x$ rounded up to the next integer, or if $x$ is negative you can just let $n_x$ be $1$.) That means $x$ is not in the set $F_n_x$, so it certainly cannot be in the intersection $bigcap_n F_n$. This is true no matter what $x$ is, so $bigcap_n F_n$ cannot contain any real numbers at all, which is to say it equals the empty set.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

It might be a little easier to understand via the contrapositive:

Let $x$ be any real number. Then we know that there exists a positive integer $n_x$ larger than $x$. (This is the so-called Archimedean property of the reals, but intuitively you can think of $n_x$ just being $x$ rounded up to the next integer, or if $x$ is negative you can just let $n_x$ be $1$.) That means $x$ is not in the set $F_n_x$, so it certainly cannot be in the intersection $bigcap_n F_n$. This is true no matter what $x$ is, so $bigcap_n F_n$ cannot contain any real numbers at all, which is to say it equals the empty set.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

$endgroup$

add a comment | 

$begingroup$

It might be a little easier to understand via the contrapositive:

Let $x$ be any real number. Then we know that there exists a positive integer $n_x$ larger than $x$. (This is the so-called Archimedean property of the reals, but intuitively you can think of $n_x$ just being $x$ rounded up to the next integer, or if $x$ is negative you can just let $n_x$ be $1$.) That means $x$ is not in the set $F_n_x$, so it certainly cannot be in the intersection $bigcap_n F_n$. This is true no matter what $x$ is, so $bigcap_n F_n$ cannot contain any real numbers at all, which is to say it equals the empty set.

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges

answered 8 hours ago

Nate EldredgeNate Eldredge

68.2k1010 gold badges8787 silver badges183183 bronze badges