MHB Why Is the Empty Set Considered Unique?

[evinda]

Hello! (Cool)

Sentence

The set, that does not contain any element, is unique.

Proof:

Let's suppose that $a,b$ are sets, so that each of these sets does not contain any element and $a \neq b$.

From the axiom: Two sets, that have the same elements, are equal., there is (without loss of generality )

$$x \in a \text{ and } x \notin b (*)$$

$a,b$ do not contain any element.

$$\forall x (x \notin a)$$
$$\forall x (x \notin b)$$

$$\forall x (x \notin a \leftrightarrow x \notin b) (**)$$

From $(*)$ and $(**)$, we have a contradiction, so the set that does not contain any element is unique.Could you explain me how we get the relation $(**)$ ?

 

[Evgeny.Makarov]

$$\forall x (x \notin a)$$
$$\forall x (x \notin b)$$

$$\forall x (x \notin a \leftrightarrow x \notin b) (**)$$

...

Could you explain me how we get the relation $(**)$ ?

In general, from $\forall x\;P(x)$ and $\forall x\;Q(x)$ we can conclude $\forall x\;(P(x)\leftrightarrow Q(x))$. Indeed, for every $x$ both sides of the equivalence $P(x)\leftrightarrow Q(x)$ are true.

I wouldn't use a proof by contradiction for this. The axiom says
\[
a=b\leftrightarrow \forall x\;(x\in a\leftrightarrow x\in b)
\]
If $a$ and $b$ are empty sets, then $x\in a$ and $x\in b$ are both false for all $x$, so $\forall x\;(x\in a\leftrightarrow x\in b)$ holds, which implies $a=b$.

 

[evinda]

In general, from $\forall x\;P(x)$ and $\forall x\;Q(x)$ we can conclude $\forall x\;(P(x)\leftrightarrow Q(x))$. Indeed, for every $x$ both sides of the equivalence $P(x)\leftrightarrow Q(x)$ are true.

I understand! (Smile)

I wouldn't use a proof by contradiction for this. The axiom says
\[
a=b\leftrightarrow \forall x\;(x\in a\leftrightarrow x\in b)
\]

If $a$ and $b$ are empty sets, then $x\in a$ and $x\in b$ are both false for all $x$, so $\forall x\;(x\in a\leftrightarrow x\in b)$ holds, which implies $a=b$.

Could you explain it further to me? (Sweating)

 

[Evgeny.Makarov]

Hmm, I am not sure what to explain. Which part is not clear?

 

[evinda]

Hmm, I am not sure what to explain. Which part is not clear?

I haven't understood why this: $\forall x (x \in a \leftrightarrow x \in b)$ holds, although $x \in a$ and $x \in b$ are both false..
Shouldn't it be $\forall x (x \notin a \leftrightarrow x \notin b)$ ? Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

$A\leftrightarrow B$ holds iff both $A$ and $B$ are true or if both of them are false.

 

[evinda]

If $a$ and $b$ are empty sets, then $x\in a$ and $x\in b$ are both false for all $x$, so $\forall x\;(x\in a\leftrightarrow x\in b)$ holds, which implies $a=b$

$A\leftrightarrow B$ holds iff both $A$ and $B$ are true or if both of them are false.

A ok.. I got it! (Nod)

Is the proof by contradiction wrong? (Thinking)

 

[Evgeny.Makarov]

Is the proof by contradiction wrong?

I wrote about it in https://driven2services.com/staging/mh/index.php?posts/58821/.

 

[evinda]

From the axiom: Two sets, that have the same elements, are equal., there is (without loss of generality )

$$x \in a \text{ and } x \notin b (*)$$

Could you explain me why the above is the contraposition of the axiom of extension? (Thinking)

 

[solakis1]

Could you explain me why the above is the contraposition of the axiom of extension? (Thinking)

By the axiom of extensionality we have:

$$a=b\Longleftrightarrow\forall x[x\in a\Longleftrightarrow x\in b]$$

THAT IMPLIES:

$$a=b\Longleftarrow\forall x[x\in a\Longleftrightarrow x\in b]$$

AND by contrapositive law we have:

$$a\neq b\Longrightarrow\neg[\forall x(x\in a\Longleftrightarrow x\in b)]$$

BUT

$$\neg[\forall x(x\in a\Longleftrightarrow x\in b)]$$ is equevalent to:

$$\exists x[ \neg((x\in a\Longrightarrow x\in b)\wedge(x\in b\longrightarrow x\in a))]$$

AND that implies ,by using D MORGAN$$ \neg(x\in a\Longrightarrow x\in b)\vee\neg(x\in b\longrightarrow x\in a)$$

Which is equivalent to:

$$(x\in a\wedge\neg x\in b)\vee(x\in b\wedge\neg x\in a)$$

 

[solakis1]

Hello! (Cool)

Sentence

The set, that does not contain any element, is unique.

Proof:

Let's suppose that $a,b$ are sets, so that each of these sets does not contain any element and $a \neq b$.

From the axiom: Two sets, that have the same elements, are equal., there is (without loss of generality )

$$x \in a \text{ and } x \notin b (*)$$

$a,b$ do not contain any element.

$$\forall x (x \notin a)$$
$$\forall x (x \notin b)$$

$$\forall x (x \notin a \leftrightarrow x \notin b) (**)$$

From $(*)$ and $(**)$, we have a contradiction, so the set that does not contain any element is unique.Could you explain me how we get the relation $(**)$ ?

A proof by contradiction is the following:

Let $$a\neq b$$

Now:

suppose $$\neg x\in a$$.......1

Since b is empty we have :

$$\forall x (\neg x\in b)$$
OR

$$ (\neg x\in b)$$........2

And by the rule of conditional proof we can conclude:

$$\neg x\in a\Longrightarrow\neg x\in b$$........3...

In the same way we infer that:

$$\neg x\in b\Longrightarrow\neg x\in a $$.................4

Now from (3) and using contrapositive we have:

$$x\in b\Longrightarrow x\in a$$................5

In same way;

$$x\in a\Longrightarrow x\in b$$.................6

FROM (5) and (6) we can conclude:

$$x\in a\longleftrightarrow x\in b$$
OR

$$\forall x[x\in a\Longleftrightarrow x\in b]$$

And by the axiom of extensionality we have: a=b a contradiction since we assumed $$a\neq b$$

Hence : a=b

There at least 5,6 different ways one can prove the uniqueness of the empty set