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All right math geeks, lay it on me. What the hell is aleph zero? Is this the right symbol for it: [itex]\aleph_0[/itex]?
- Warren
- Warren
Originally posted by chroot
All right math geeks, lay it on me. What the hell is aleph zero? Is this the right symbol for it: [itex]\aleph_0[/itex]?
Originally posted by chroot
All right math geeks, lay it on me. What the hell is aleph zero? Is this the right symbol for it: [itex]\aleph_0[/itex]?
- Warren
Originally posted by chroot
So... a set with cardinality [itex]\aleph_0[/itex] has countably many elements?
What is the cardinality of the set of real numbers? They are uncountably infinite, right?
Also, I've seen people use [itex]\aleph_0[/itex] like a number -- they'' even say stuff like [itex]2^{\aleph_0}[/itex]. This just doesn't make any sense to me. Is it a number? If not, what is it?
- Warren
Originally posted by chroot
So... a set with cardinality [itex]\aleph_0[/itex] has countably many elements?
What is the cardinality of the set of real numbers? They are uncountably infinite, right?
Also, I've seen people use [itex]\aleph_0[/itex] like a number -- they'' even say stuff like [itex]2^{\aleph_0}[/itex]. This just doesn't make any sense to me. Is it a number? If not, what is it?
- Warren
{
{(0,A),(1,A)}
{(0,A),(1,B)}
{(0,B),(1,A)}
{(0,B),(1,B)}
}
Originally posted by Hurkyl
Minor correction; [itex]A^B[/itex] is the set of all functions from [itex]B[/itex] to [itex]A[/itex], and exponentiation for cardinal numbers is defined as [itex]|A|^{|B|}= |A^B| [/itex].
Originally posted by chroot
Okay so [itex]\aleph_0 = | \mathbb{Z} | = | \mathbb{N} | = | \mathbb{Q} |[/itex] and [itex]\aleph_1 = | \mathbb{R} |[/itex].
Is it acceptable to say that [itex]\aleph_1 > \aleph_0[/itex]? Or that the cardinality of the reals is larger than the cardinality of the integers?
I'm not sure I understand where the [itex]\mathfrak{c}[/itex] came from if [itex]\aleph_1 \equiv \mathfrak{c}[/itex].
Is there an [itex]\aleph_2[/itex], ad infinitum? This all seems funny to me, that these cardinal numbers obey different sorts of rules than normal numbers. I haven't gotten my head around it yet.
- Warren
Originally posted by chroot
Is it acceptable to say that [itex]\aleph_1 > \aleph_0[/itex]?
I'm not sure I understand where the [itex]\mathfrak{c}[/itex] came from
Is there an [itex]\aleph_2[/itex], ad infinitum?
Originally posted by Hurkyl
the class of cardinal numbers is "too big" to fit in a set.
Can you expand a little on this?
*mumbles* mommy.. mommy.. make it stop.Originally posted by Hurkyl
Therefore the set of all cardinal numbers cannot exist.
Originally posted by NateTG
Consider this:
[tex]|2^A| > |A|[/tex]
is strict for
[tex]A \neq 0[/tex]
Let [tex]G[/tex] be a mapping [tex]A \rightarrow \{0,1\}^A [/tex]. Then for every [tex]a \in A[/tex], [tex]G(a)[/tex] is a function [tex]A \rightarrow \{0,1\}[/tex]
Now, construct [tex]f:A \rightarrow \{0,1\}[/tex] in the following way:
[tex]f(a)= 1 [/tex] if [tex]G(a)(a)=0[/tex] and [tex]0[/tex] otherwise.
Clearly [tex]f[/tex] is not in the range of [tex]G[/tex] since [tex]G(a)(a) \neq f(a) \forall a \in A[/tex]
Therefore there are no surjective mappings [tex]A \rightarrow \{0,1\}^A [/tex], and no bijections can exist.
Proving the other direction is easy:
[tex]G(a)(b)=1 \iff a=b[/tex] is an injective function.
This proves that there are 'infinitely large' infinities.
Yep - but this is the grown-up versionOriginally posted by master_coda
Isn't that just Cantor's diagonal method?
Originally posted by chroot
What the hell is aleph zero?
Originally posted by uart
Question ? Some people above are referring to [tex] \aleph_1[/tex] as being equal [tex] \aleph_0^{\aleph_0}[/tex]. But why isn't [tex] \aleph_1[/tex] equal to [tex] 2^{\aleph_0}[/tex], since I've seen it shown that this is the next cardinal greater than [tex] \aleph_0[/tex] ?
Originally posted by master_coda
It hasn't been shown that [itex]2^{\aleph_0}[/itex] is the next cardinal greater than [itex]\aleph_0[/itex]. It can't be shown - since [itex]2^{\aleph_0}=\mathfrak{c}[/itex], the idea that [itex]2^{\aleph_0}=\aleph_1[/itex] is just a restatement of the continuum hypothesis.
Originally posted by uart
But since [tex]2^{\aleph_0}[/tex] can't be put into a 1-1 relation with the natural numbers then doesn't that mean that [tex]2^{\aleph_0}[/tex] is larger than [tex]\aleph_0[/tex] ? And if that is the case then why do you need to go all the way to [tex]\aleph_0^{\aleph_0}[/tex] to find the next thing bigger when [tex]2^{\aleph_0}[/tex] is bigger already ?
You have a right to be confused! Except for uart'sBut since can't be put into a 1-1 relation with the natural numbers then doesn't that mean that is larger than [tex]\aleph_0^{\aleph_0}[/tex] ? And if that is the case then why do you need to go all the way to to find the next thing bigger when is bigger already ?
I have never seen anyone refer to [tex]\aleph_0^{\aleph_0}[/tex]!Some people above are referring to [tex]\aleph_0^{\aleph_0}[/tex] as being equal .
Originally posted by HallsofIvy
I have never seen anyone refer to [tex]\aleph_0^{\aleph_0}[/tex]!
Where did you get that? I was under the impression [itex]{2}^{\aleph_0}= \mathfrak{c}[/itex] and that the assertion that this was equal to [itex]{\aleph_1}[/itex] was the "continuum hypothesis" but I will say again that I have never seen [itex]{\aleph_0}^{\aleph_0}[/itex] beforeSince [itex]{\aleph_0}^{\aleph_0}=\mathfrak{c}[/itex]
Originally posted by HallsofIvy
Where did you get that? I was under the impression [itex]{2}^{\aleph_0}= \mathfrak{c}[/itex] and that the assertion that this was equal to [itex]{\aleph_1}[/itex] was the "continuum hypothesis" but I will say again that I have never seen [itex]{\aleph_0}^{\aleph_0}[/itex] before