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

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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

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asked 8 hours ago

mavaviljmavavilj

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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

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share|cite|improve this question

asked 8 hours ago

mavaviljmavavilj

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asked 8 hours ago

mavaviljmavavilj

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asked 8 hours ago

mavaviljmavavilj

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4 Answers
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5

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

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answered 8 hours ago

BernardBernard

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3

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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.

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answered 8 hours ago

pre-kidneypre-kidney

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1

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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

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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

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5

$begingroup$

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

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add a comment | 

5

$begingroup$

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

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$endgroup$

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

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$endgroup$

share|cite|improve this answer

answered 8 hours ago

BernardBernard

133k77 gold badges4343 silver badges126126 bronze badges

answered 8 hours ago

BernardBernard

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answered 8 hours ago

BernardBernard

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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

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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

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$endgroup$

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

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$endgroup$

share|cite|improve this answer

answered 8 hours ago

pre-kidneypre-kidney

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answered 8 hours ago

pre-kidneypre-kidney

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answered 8 hours ago

pre-kidneypre-kidney

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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

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$endgroup$

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

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$endgroup$

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