# Numerically Equivalent Sets

#### novice

For infinite sets $$\displaystyle A$$ and $$\displaystyle B$$ to be numerically equivalent. Is bijective function

$$\displaystyle f:A \rightarrow B$$

the only requirement that needs satisfying?

#### gmatt

For infinite sets $$\displaystyle A$$ and $$\displaystyle B$$ to be numerically equivalent. Is bijective function

$$\displaystyle f:A \rightarrow B$$

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.)

• novice

#### Drexel28

MHF Hall of Honor
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 $$\displaystyle \text{card }A=\alpha,\text{card }B=\beta$$ then $$\displaystyle \alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B$$

• novice

#### novice

If $$\displaystyle \text{card }A=\alpha,\text{card }B=\beta$$ then $$\displaystyle \alpha\leqslant\beta\Leftrightarrow \exists f:A\overset{\text{injection}}{\longrightarrow}B$$
Are you thinking of $$\displaystyle A$$ and $$\displaystyle B$$ being countable?
In my question, $$\displaystyle A$$ and $$\displaystyle B$$ 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 $$\displaystyle A$$ nor $$\displaystyle B$$ is denumerable, except there is a bijective function from A to B.

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

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

is defined by $$\displaystyle f(x)=x$$

Since we know that $$\displaystyle f$$ is bijective, despite the fact that $$\displaystyle A$$ and $$\displaystyle B$$ being uncountable, is it sufficient to conclude that $$\displaystyle |A|=|B|$$?

#### novice

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.

#### Defunkt

MHF Hall of Honor
If I recall correctly, the definition is that $$\displaystyle |A| = |B| \text{ if } \exists f:A \overset{\text{bijection}} {\longrightarrow} B$$, and $$\displaystyle |A| \leq |B| \text{ if } \exists f:A \overset{\text{injection}} {\longrightarrow} B$$, as drexel has pointed out.

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

• novice

#### novice

If I recall correctly, the definition is that $$\displaystyle |A| = |B| \text{ if } \exists f:A \overset{\text{bijection}} {\longrightarrow} B$$, and $$\displaystyle |A| \leq |B| \text{ if } \exists f:A \overset{\text{injection}} {\longrightarrow} B$$, 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 $$\displaystyle f$$ is injective, the $$\displaystyle |A| \leq |B|$$, but I couldn't see how that automatically translate to $$\displaystyle f$$ being bijective.

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

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

Yah?

#### Swlabr

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 $$\displaystyle f$$ is injective, the $$\displaystyle |A| \leq |B|$$, but I couldn't see how that automatically translate to $$\displaystyle f$$ being bijective.

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

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

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

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

• novice

#### novice

... you don't need to go the whole hog and find a bijection!
(Rofl)

#### novice

Friends, I thought I was done, but something is troubling me when I made $$\displaystyle A=\varnothing$$ and $$\displaystyle B=\varnothing$$.

Cleary $$\displaystyle \varnothing$$ is finite, and we all know that $$\displaystyle |\varnothing|=|\varnothing|.$$ Since there isn't any element to count nor is there a function from $$\displaystyle \varnothing$$ to $$\displaystyle \varnothing$$, how do you prove that |A|=|B|?