Is the cardinality of the set of points on the circumference of a circle equal to the cardinality of the set...
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Is the cardinality of the set of points on the circumference of a circle equal to the cardinality of the set of points on the interior of a circle? They're both infinite, but are they the same aleph?
general-topology geometry elementary-set-theory
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
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add a comment |
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Is the cardinality of the set of points on the circumference of a circle equal to the cardinality of the set of points on the interior of a circle? They're both infinite, but are they the same aleph?
general-topology geometry elementary-set-theory
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
$endgroup$
$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
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– David
Dec 14 '18 at 3:06
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Is this a general-topology problem?
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– YuiTo Cheng
Dec 14 '18 at 16:55
add a comment |
$begingroup$
Is the cardinality of the set of points on the circumference of a circle equal to the cardinality of the set of points on the interior of a circle? They're both infinite, but are they the same aleph?
general-topology geometry elementary-set-theory
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
$endgroup$
Is the cardinality of the set of points on the circumference of a circle equal to the cardinality of the set of points on the interior of a circle? They're both infinite, but are they the same aleph?
general-topology geometry elementary-set-theory
general-topology geometry elementary-set-theory
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
edited Dec 14 '18 at 3:09
spaceisdarkgreen
32.8k21753
32.8k21753
asked Dec 14 '18 at 2:55
DavidDavid
1506
asked Dec 14 '18 at 2:55
DavidDavid
1506
asked Dec 14 '18 at 2:55
DavidDavid
1506
1506
$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
$endgroup$
– David
Dec 14 '18 at 3:06
$begingroup$
Is this a general-topology problem?
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 16:55
add a comment |
$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
$endgroup$
– David
Dec 14 '18 at 3:06
$begingroup$
Is this a general-topology problem?
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 16:55
$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
$endgroup$
– David
Dec 14 '18 at 3:06
$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
$endgroup$
– David
Dec 14 '18 at 3:06
$begingroup$
Is this a general-topology problem?
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 16:55
$begingroup$
Is this a general-topology problem?
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 16:55
add a comment |
1 Answer
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Look at it this way: you're comparing the cardinality of the points on a line, with those in an area of the plane.
Obviously both are finite (finite length and area I mean, not cardinality), but you know that the interval $(0,1)$ has the same cardinality as that of the entire number line $mathbb{R}$ - both having cardinality of the continuum, $c$. It should then come to no surprise - someone could probably flesh out the details - that the cardinality of points enclosed by a circle in the plane with nonzero area would have the same cardinality as the plane itself.
In this sense, then, the circle's circumference is analogous to a finite interval of $mathbb{R}$, and similarly the circle encloses a finite area of $mathbb{R} times mathbb{R}$. The cardinalities we're concerned with then, in the end, are the cardinalities of $mathbb{R}$ and $mathbb{R} times mathbb{R}$, right?
For a Cartesian product, we note that $|A times B | = |A| cdot |B|$. Thus,
$$|mathbb{R} times mathbb{R}| = |mathbb{R}| cdot |mathbb{R}| = c cdot c = c = |mathbb{R}|$$
(Counterintuitive as $c cdot c = c$ would be, it's true. In fact, in general, $| mathbb{R}^n | = c$ for all $n in mathbb{N}$. That's another detail someone else more qualified might want to hash out if you're confused though.)
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
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add a comment |
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1 Answer
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1 Answer
1
active
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active
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$begingroup$
Look at it this way: you're comparing the cardinality of the points on a line, with those in an area of the plane.
Obviously both are finite (finite length and area I mean, not cardinality), but you know that the interval $(0,1)$ has the same cardinality as that of the entire number line $mathbb{R}$ - both having cardinality of the continuum, $c$. It should then come to no surprise - someone could probably flesh out the details - that the cardinality of points enclosed by a circle in the plane with nonzero area would have the same cardinality as the plane itself.
In this sense, then, the circle's circumference is analogous to a finite interval of $mathbb{R}$, and similarly the circle encloses a finite area of $mathbb{R} times mathbb{R}$. The cardinalities we're concerned with then, in the end, are the cardinalities of $mathbb{R}$ and $mathbb{R} times mathbb{R}$, right?
For a Cartesian product, we note that $|A times B | = |A| cdot |B|$. Thus,
$$|mathbb{R} times mathbb{R}| = |mathbb{R}| cdot |mathbb{R}| = c cdot c = c = |mathbb{R}|$$
(Counterintuitive as $c cdot c = c$ would be, it's true. In fact, in general, $| mathbb{R}^n | = c$ for all $n in mathbb{N}$. That's another detail someone else more qualified might want to hash out if you're confused though.)
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
$endgroup$
add a comment |
$begingroup$
Look at it this way: you're comparing the cardinality of the points on a line, with those in an area of the plane.
Obviously both are finite (finite length and area I mean, not cardinality), but you know that the interval $(0,1)$ has the same cardinality as that of the entire number line $mathbb{R}$ - both having cardinality of the continuum, $c$. It should then come to no surprise - someone could probably flesh out the details - that the cardinality of points enclosed by a circle in the plane with nonzero area would have the same cardinality as the plane itself.
In this sense, then, the circle's circumference is analogous to a finite interval of $mathbb{R}$, and similarly the circle encloses a finite area of $mathbb{R} times mathbb{R}$. The cardinalities we're concerned with then, in the end, are the cardinalities of $mathbb{R}$ and $mathbb{R} times mathbb{R}$, right?
For a Cartesian product, we note that $|A times B | = |A| cdot |B|$. Thus,
$$|mathbb{R} times mathbb{R}| = |mathbb{R}| cdot |mathbb{R}| = c cdot c = c = |mathbb{R}|$$
(Counterintuitive as $c cdot c = c$ would be, it's true. In fact, in general, $| mathbb{R}^n | = c$ for all $n in mathbb{N}$. That's another detail someone else more qualified might want to hash out if you're confused though.)
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
$endgroup$
add a comment |
$begingroup$
Look at it this way: you're comparing the cardinality of the points on a line, with those in an area of the plane.
Obviously both are finite (finite length and area I mean, not cardinality), but you know that the interval $(0,1)$ has the same cardinality as that of the entire number line $mathbb{R}$ - both having cardinality of the continuum, $c$. It should then come to no surprise - someone could probably flesh out the details - that the cardinality of points enclosed by a circle in the plane with nonzero area would have the same cardinality as the plane itself.
In this sense, then, the circle's circumference is analogous to a finite interval of $mathbb{R}$, and similarly the circle encloses a finite area of $mathbb{R} times mathbb{R}$. The cardinalities we're concerned with then, in the end, are the cardinalities of $mathbb{R}$ and $mathbb{R} times mathbb{R}$, right?
For a Cartesian product, we note that $|A times B | = |A| cdot |B|$. Thus,
$$|mathbb{R} times mathbb{R}| = |mathbb{R}| cdot |mathbb{R}| = c cdot c = c = |mathbb{R}|$$
(Counterintuitive as $c cdot c = c$ would be, it's true. In fact, in general, $| mathbb{R}^n | = c$ for all $n in mathbb{N}$. That's another detail someone else more qualified might want to hash out if you're confused though.)
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
$endgroup$
Look at it this way: you're comparing the cardinality of the points on a line, with those in an area of the plane.
Obviously both are finite (finite length and area I mean, not cardinality), but you know that the interval $(0,1)$ has the same cardinality as that of the entire number line $mathbb{R}$ - both having cardinality of the continuum, $c$. It should then come to no surprise - someone could probably flesh out the details - that the cardinality of points enclosed by a circle in the plane with nonzero area would have the same cardinality as the plane itself.
In this sense, then, the circle's circumference is analogous to a finite interval of $mathbb{R}$, and similarly the circle encloses a finite area of $mathbb{R} times mathbb{R}$. The cardinalities we're concerned with then, in the end, are the cardinalities of $mathbb{R}$ and $mathbb{R} times mathbb{R}$, right?
For a Cartesian product, we note that $|A times B | = |A| cdot |B|$. Thus,
$$|mathbb{R} times mathbb{R}| = |mathbb{R}| cdot |mathbb{R}| = c cdot c = c = |mathbb{R}|$$
(Counterintuitive as $c cdot c = c$ would be, it's true. In fact, in general, $| mathbb{R}^n | = c$ for all $n in mathbb{N}$. That's another detail someone else more qualified might want to hash out if you're confused though.)
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
answered Dec 14 '18 at 3:09
Eevee TrainerEevee Trainer
5,9771936
5,9771936
add a comment |
add a comment |
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$begingroup$
Why is the answer yes, I would like to have knowledge of this. Carnal or otherwise.
$endgroup$
– David
Dec 14 '18 at 3:06
$begingroup$
Is this a general-topology problem?
$endgroup$
– YuiTo Cheng
Dec 14 '18 at 16:55