<p>Our goal is to prove that $\aleph_1 &lt; 2^{\aleph_0}$ in the model $M[G]$, and therefore the Continuum Hypothesis is not true in $M[G]$. This is equivalent to saying that $\aleph_2 \leq 2^{\aleph_0}$. So we need to construct a model $M[G]$ such that there is an injective map $f$ from $\aleph_2$ to $2^{\aleph_0}$ in $M[G]$. Note that each element of  $2^{\aleph_0}$ is a function from $\aleph_0$ to $\{0,1\}$ (by definition). Thus $f(x)$ (the value of $f$ at point $x$) is actually a function that maps a natural number to $\{0,1\}$. In the proof, we treat $f$ as a function of two parameters
$$h(x,y) = f(x)(y)$$
where $x\in \aleph_2$ and $y\in \aleph_0$. This is the reason why we want to prove that there exist a function $h$ from $\aleph_2 \times \aleph_0$ to $\{0,1\}$ satisfying certain properties. </p>

<p>The choice of $\aleph_2$ here is not arbitrary. Say, if we wanted to prove that $2^{\aleph_0} &gt; \aleph_{19}$, we would consider functions from $\aleph_{20} \times \aleph_0$ to $\{0,1\}$.</p>

<p>BTW, $\aleph_2^M$ is always an element of $M$ and $M[G]$ (for any generic set $G$); on the other hand $\aleph_2$ is usually neither an element of $M$ nor $M[G]$, since $M$ is a <em>countable</em> transitive model of ZFC (or more precisely is a countable model of a large finite fragment of ZFC). So I'm not sure what you mean in the last paragraph.</p>

CC BY-SA 3.0
2012-12-03T17:14:28.700Z
Great explanation, Yury.  As for the last paragraph, you are right, I skipped the qualification of $\aleph_2$ as the $\aleph_2$ within M (i.e. 'relativized' to M).
djkern
CC BY-SA 3.0
2012-12-05T23:36:08.813Z
In some sense the choice of $\aleph_2$ _is_ arbitrary, isn't it?  i.e., there's no special property of $\aleph_2$ that's being used for the argument...
Steven Stadnicki
CC BY-SA 3.0
2012-12-06T00:00:03.840Z
@StevenStadnicki: I agree. We could choose any cardinal $\apha > \aleph_1$ instead and prove that $2^{\aleph_0} \geq \alpha$. That would also show that CH doesn't hold in our model. But at least, the choice of $\aleph_2$ in some sense is *very natural*.
Yury
CC BY-SA 3.0
2012-12-03T17:00:08.667Z
2012-12-03T17:24:35.643Z
2012-12-03T17:24:35.643Z
Yury
<p>I've read a few different presentations of Cohen's proof. All of them (that I've seen) eventually make a move where a Cartesian product (call it CP) between the (M-form of) $\aleph_2$ and $\aleph_0$ into {1, 0} is imagined. From what I gather, whether $CP \in M$ is what determines whether $\neg$CH holds in M or not such that if $CP \in M$ then $\neg$CH.</p>

<p>Anyway, my question is: Why does this product, CP, play this role? How does it show us that $\aleph_2 \in M$ (the 'relativized' form of $\aleph_2$, not the $\aleph_2 \in V$)? Could not some other set-theoretical object play the same role?</p>

CC BY-SA 3.0
2012-12-03T16:53:23.613Z
You could probably get a faster answer at http://mathoverflow.net/ . This is more of a mathematical question than it is a computer science one (not that I would mind a strong logic community in this forum).
cody
CC BY-SA 3.0
2012-12-03T18:45:27.450Z
Didn't take that long :)
Suresh Venkat
CC BY-SA 3.0
2012-12-06T00:53:42.080Z
That said, it still might make sense to migrate this Q over to math.SE since it's a much more natural fit there than in TCS-world...
Steven Stadnicki
<lo.logic><set-theory>
Trying understand a move in Cohen's proof of the independence of the continuum hypothesis
djkern
# Trying understand a move in Cohen's proof of the independence of the continuum hypothesis

(Asked by: username_0 on 2012-12-03)

I've read a few different presentations of Cohen's proof. All of them (that I've seen) eventually make a move where a Cartesian product (call it CP) between the (M-form of) $\aleph\_2$ and $\aleph\_0$ into {1, 0} is imagined. From what I gather, whether $CP \in M$ is what determines whether $\neg$CH holds in M or not such that if $CP \in M$ then $\neg$CH.

Anyway, my question is: Why does this product, CP, play this role? How does it show us that $\aleph\_2 \in M$ (the 'relativized' form of $\aleph\_2$, not the $\aleph\_2 \in V$)? Could not some other set-theoretical object play the same role?

**Tags:** lo.logic, set-theory

> **Comments:**
> **username_1**: You could probably get a faster answer at http://mathoverflow.net/ . This is more of a mathematical question than it is a computer science one (not that I would mind a strong logic community in this forum).  
> **username_2**: Didn't take that long :)  
> **username_3**: That said, it still might make sense to migrate this Q over to math.SE since it's a much more natural fit there than in TCS-world...  

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## 1 answers

### Answer by username_4 on 2012-12-03. Score: 14 | Accepted answer

Our goal is to prove that $\aleph\_1 < 2^{\aleph\_0}$ in the model $M[G]$, and therefore the Continuum Hypothesis is not true in $M[G]$. This is equivalent to saying that $\aleph\_2 \leq 2^{\aleph\_0}$. So we need to construct a model $M[G]$ such that there is an injective map $f$ from $\aleph\_2$ to $2^{\aleph\_0}$ in $M[G]$. Note that each element of $2^{\aleph\_0}$ is a function from $\aleph\_0$ to $\{0,1\}$ (by definition). Thus $f(x)$ (the value of $f$ at point $x$) is actually a function that maps a natural number to $\{0,1\}$. In the proof, we treat $f$ as a function of two parameters
$$h(x,y) = f(x)(y)$$
where $x\in \aleph\_2$ and $y\in \aleph\_0$. This is the reason why we want to prove that there exist a function $h$ from $\aleph\_2 \times \aleph\_0$ to $\{0,1\}$ satisfying certain properties.

The choice of $\aleph\_2$ here is not arbitrary. Say, if we wanted to prove that $2^{\aleph\_0} > \aleph\_{19}$, we would consider functions from $\aleph\_{20} \times \aleph\_0$ to $\{0,1\}$.

BTW, $\aleph\_2^M$ is always an element of $M$ and $M[G]$ (for any generic set $G$); on the other hand $\aleph\_2$ is usually neither an element of $M$ nor $M[G]$, since $M$ is a *countable* transitive model of ZFC (or more precisely is a countable model of a large finite fragment of ZFC). So I'm not sure what you mean in the last paragraph.

> **Comments:**
> **username_0**: Great explanation, username_4.  As for the last paragraph, you are right, I skipped the qualification of $\aleph_2$ as the $\aleph_2$ within M (i.e. 'relativized' to M).  
> **username_3**: In some sense the choice of $\aleph_2$ _is_ arbitrary, isn't it?  i.e., there's no special property of $\aleph_2$ that's being used for the argument...  
> **username_4**: @StevenStadnicki: I agree. We could choose any cardinal $\apha > \aleph_1$ instead and prove that $2^{\aleph_0} \geq \alpha$. That would also show that CH doesn't hold in our model. But at least, the choice of $\aleph_2$ in some sense is *very natural*.  

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