# Numerically Equivalent Sets

• May 11th 2010, 07:18 PM
novice
Numerically Equivalent Sets
For infinite sets $A$ and $B$ to be numerically equivalent. Is bijective function

$f:A \rightarrow B$

the only requirement that needs satisfying?
• May 11th 2010, 08:30 PM
gmatt
Quote:

Originally Posted by novice
For infinite sets $A$ and $B$ to be numerically equivalent. Is bijective function

$f:A \rightarrow B$

the only requirement that needs satisfying?

I'm not sure what you mean by numerically equivalent. One way of defining two sets to have the same cardinality if and only if there exists a bijection between the two.

There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.

(Note that when I say if and only if its merely a logical statement used in every definition, sometimes implicitly.)
• May 11th 2010, 08:40 PM
Drexel28
Quote:

Originally Posted by gmatt

There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.

There is nothing new here. If $\text{card }A=\alpha,\text{card }B=\beta$ then $\alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B$
• May 12th 2010, 04:11 AM
novice
Quote:

Originally Posted by Drexel28
If $\text{card }A=\alpha,\text{card }B=\beta$ then $\alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B$

Are you thinking of $A$ and $B$ being countable?
In my question, $A$ and $B$ are infinite.

I know this theorem: An infinite subset of a denumerable set is denumerable.

But I am tweaking it for a different scenario, where I know that neither $A$ nor $B$ is denumerable, except there is a bijective function from A to B.

To make it simple. Suppose that $A = \mathbb{R}$ and $B=\mathbb{R}$

$f:A \rightarrow B$ or equivalently $f:\mathbb{R} \rightarrow \mathbb{R}$

is defined by $f(x)=x$

Since we know that $f$ is bijective, despite the fact that $A$ and $B$ being uncountable, is it sufficient to conclude that $|A|=|B|$?
• May 12th 2010, 04:14 AM
novice
Quote:

Originally Posted by gmatt
I'm not sure what you mean by numerically equivalent. One way of defining two sets to have the same cardinality if and only if there exists a bijection between the two.

There is apparently another way of comparing set cardinalities, with cardinal numbers but I never learned anything on the topic and someone more knowledgeable on the matter will have to discuss it.

(Note that when I say if and only if its merely a logical statement used in every definition, sometimes implicitly.)

Numerically equivalent set by definition is |A|=|B| where the sets are infinite. Please see more details in the question I put to Drexel.
• May 12th 2010, 05:12 AM
Defunkt
If I recall correctly, the definition is that $|A| = |B| \text{ if } \exists f:A \overset{\text{bijection}} {\longrightarrow} B$, and $|A| \leq |B| \text{ if } \exists f:A \overset{\text{injection}} {\longrightarrow} B$, as drexel has pointed out.

(This is the definition for any A,B - not just countable sets.)
• May 12th 2010, 06:25 AM
novice
Quote:

Originally Posted by Defunkt
If I recall correctly, the definition is that $|A| = |B| \text{ if } \exists f:A \overset{\text{bijection}} {\longrightarrow} B$, and $|A| \leq |B| \text{ if } \exists f:A \overset{\text{injection}} {\longrightarrow} B$, as drexel has pointed out.

(This is the definition for any A,B - not just countable sets.)

Well, you have made it very clear for me to understand.

Drexel runs at hypersonic speed when I am still crawling. I only knew that if $f$ is injective, the $|A| \leq |B|$, but I couldn't see how that automatically translate to $f$ being bijective.

Now, I think you mean if $f$ is bijective, then $f:A\rightarrow B \Rightarrow |A|\leq |B|$, and if $f$ is surjective, then $|A|\geq |B|$. Then only after putting those two together,

$|A|\leq |B|$ and $|A|\geq |B| \Rightarrow |A|=|B|$ .

Yah?
• May 12th 2010, 06:32 AM
Swlabr
Quote:

Originally Posted by novice
Well, you have made it very clear for me to understand.

Drexel runs at hypersonic speed when I am still crawling. I only knew that if $f$ is injective, the $|A| \leq |B|$, but I couldn't see how that automatically translate to $f$ being bijective.

Now, I think you mean if $f$ is bijective, then $f:A\rightarrow B \Rightarrow |A|\leq |B|$, and if $f$ is surjective, then $|A|\geq |B|$. Then only after putting those two together,

$|A|\leq |B|$ and $|A|\geq |B| \Rightarrow |A|=|B|$ .

Yah?

Yes, if $f: A \rightarrow B$ surjective then this implies $A \geq B$. However, the general technique is that if $f_1:A \rightarrow B$ is an injection and $f_2:B \rightarrow A$ is also an injection, then $|A| \leq |B|$ and $|B| \leq |A|$ so $|A| = |B|$.

For example, there clearly exists an injection $f: \mathbb{N} \rightarrow \mathbb{Q}$, this is just $a \mapsto a$. So to prove that $|\mathbb{N}| = |\mathbb{Q}|$ you would just need to find an injection from $\mathbb{Q}$ to $\mathbb{N}$, you don't need to go the whole hog and find a bijection!
• May 12th 2010, 06:36 AM
novice
Quote:

Originally Posted by Swlabr
... you don't need to go the whole hog and find a bijection!

(Rofl)
• May 12th 2010, 07:55 AM
novice
Friends, I thought I was done, but something is troubling me when I made $A=\varnothing$ and $B=\varnothing$.

Cleary $\varnothing$ is finite, and we all know that $|\varnothing|=|\varnothing|.$ Since there isn't any element to count nor is there a function from $\varnothing$ to $\varnothing$, how do you prove that |A|=|B|?
• May 12th 2010, 07:59 AM
MoeBlee
Quote:

Originally Posted by novice
nor is there a function from $\varnothing$ to $\varnothing$

There is a bijection from the empty set onto the empty set. The empty set itself is a bijection from the empty set onto the empty set.

Moreover, if A = the empty set, and B = the empty set, then you don't need set theory to prove |A| = |B|, since it follows by identity theory alone (A = the empty set = B, so A = B, so |A| = |B|).
• May 12th 2010, 08:05 AM
novice
Quote:

Originally Posted by MoeBlee
There is a bijection from the empty set onto the empty set. The empty set itself is a bijection from the empty set onto the empty set.

Moreover, if B = the empty set, and A = the empty set, then you don't need set theory to prove |A| = |B|, since it follows by identity theory alone.

Oh! You are absolutely right. However dumb it was, I am so glad I asked.
• May 12th 2010, 08:10 AM
MoeBlee
By the way, I think in some of the discussion above it was taken as implicit that if there is a surjection from A onto B then there is an injection from B into A. However, just to note, that requires the axiom of choice.

The salient principles here that do not require the axiom of choice:

By definition, A and B are equinumerous if and only if there is a bijection from A onto B.

And, if there is a bijection from A onto B, then there is a bijection from B onto A (just take the inverse of the bijection from A onto B).

And, if there is an injection from A into B and there is an injection from B into A, then there is a bijection from A onto B. And, such a bijection can be recovered from the two given injections (by following any of the constructive proofs of Schroder-Bernstein).
• May 12th 2010, 04:58 PM
Drexel28
Quote:

Originally Posted by MoeBlee
By the way, I think in some of the discussion above it was taken as implicit that if there is a surjection from A onto B then there is an injection from B into A. However, just to note, that requires the axiom of choice.

The salient principles here that do not require the axiom of choice:

By definition, A and B are equinumerous if and only if there is a bijection from A onto B.

And, if there is a bijection from A onto B, then there is a bijection from B onto A (just take the inverse of the bijection from A onto B).

And, if there is an injection from A into B and there is an injection from B into A, then there is a bijection from A onto B. And, such a bijection can be recovered from the two given injections (by following any of the constructive proofs of Schroder-Bernstein).

I would like to remark that just like the above remark there is another thing which needs to be proven using Zorn's Lemma (AKA AOC).

Theorem: $\text{has a smaller cardinal number}\overset{\Delta}{=}\text{ }\leqslant$ defines a linear ordering on the set of all cardinal numbers.

Proof: The only thing that is unclear is the linearity of $\leqslant$. So, we must prove that given sets $A,B$ it is true that $\exists f:A\overset{\text{injection}}{\longrightarrow}B\te xt{ or }\exists g:B\overset{\text{injection}}{\longrightarrow}A$. If either $A$ or $B$ is empty the result is clear, so assume not. Fix $a_0\in A$ and $b_0\in B$ and define $f_0:\{a_0\}\to\{b_0\}$ in the only possible way, clearly this is an injection. Let $\Omega=\left\{(E,G,f):\{a_0\}\subseteq E,\text{ }\{b_0\}\subseteq G,\text{ }f:E\overset{\text{bijection}}{\longrightarrow}G,\ text{ } x\overset{f}{\mapsto}y\right\}$. Define the ordering $\overset{\Omega}{\leqslant}$ on $\Omega$ by $(E,G,f)\overset{\Omega}{\leqslant}\left(E',G',f'\r ight)\Leftrightarrow E\subseteq E',\text{ }G\subseteq G',\text{ }f'\mid_{E}=f$. This clearly is a partial ordering. We now must prove that every chain $\left\{\left(E_\alpha,G_\alpha,f_\alpha\right)\rig ht\}_{\alpha\in\mathcal{A}}$ has an upper bound. But, it is clear (I hope) that $\left(\bigcup_{\alpha\in\mathcal{A}}E_\alpha,\bigc up_{\alpha\in\mathcal{A}}G_\alpha,F\right)\in\Omeg a$ where $F:\bigcup_{\alpha\in\mathcal{A}}E_\alpha\to\bigcup _{\alpha\in\mathcal{A}}G_\alpha$ is defined in the obvious way (@novice there's an exercise, figure it out how to say that in a rigorous manner). Also, once you have defined $F$ it will be clear that $\left(\bigcup_{\alpha\in\mathcal{A}}E_\alpha,\bigc up_{\alpha\in\mathcal{A}}G_\alpha,F\right)$ is an upper bound for $\left\{\left(E_\alpha,G_\alpha,f_\alpha\right)\rig ht\}_{\alpha\in\mathcal{A}}$ from where it follows from Zorn's Lemma that $\left(\Omega,\overset{\Omega}{\leqslant}\right)$ must have a maximal element $\left(E_\mathfrak{M},G_\mathfrak{M},f_\mathfrak{M} \right)$. Now, suppose that $E_\mathfrak{M}\ne A\text{ and }G_\mathfrak{M}\ne B$. Then, choosing $x\in A-E_{\mathfrak{M}}\text{ and }y\in B-G_{\mathfrak{M}}$ we may define $\alpha:E_{\mathfrak{M}}\cup\{x\}\to G_{\mathfrak{M}}\cup\{y\}$ by $\alpha\mid_{E_\mathfrak{M}}=f_{\mathfrak{M}}$ and $x\overset{\alpha}{\mapsto}y$ and then of course $\left(E_{\mathfrak{M}}\cup\{x\},G_{\mathfrak{M}}\c up\{y\},\alpha\right)$ would contradict the maximality of $\left(E_{\mathfrak{M}},G_{\mathfrak{M}},f_{\mathfr ak{M}}\right)$.

Thus, there exists a bijection from one of $A$ or $B$ into a subset of the other, and the conclusion follows. $\blacksquare$

I don't think I said that very eloquently, but oh well.

In essence all your doing is pairing off elements until you run out.