MHB Is A the Empty Set When A is a Subset of the Empty Set?

[evinda]

Hello! (Wave)

I want to show that $A \subset \varnothing \rightarrow A=\varnothing$.

That's what I thought:

$$A \subset \varnothing \text{ means that :}$$
$$\forall x (x \in A \rightarrow x \in \varnothing)$$

Since, there is no $x$, such that $x \in \varnothing$, there is no $x$, such that $x \in A$.

Therefore, $A$ is the empty set.

Could you tell me if it is right? (Thinking)

 

[Siron]

Well, we need to prove that if $ A \subset \emptyset$ it follows that $A = \emptyset$.

Proof
Suppose $A \neq \emptyset$ then we can find an $x \in A$ such that $x \notin \emptyset$. Since $A \subset \emptyset$ it follows that $x \notin A$ hence we have a contradiction. Therefore our assumption was false and thus $A = \emptyset$.

 

[Evgeny.Makarov]

$$A \subset \varnothing \text{ means that :}$$
$$\forall x (x \in A \rightarrow x \in \varnothing)$$

Since, there is no $x$, such that $x \in \varnothing$, there is no $x$, such that $x \in A$.

Therefore, $A$ is the empty set.

Could you tell me if it is right?

Yes, you are right.

 

[evinda]

Well, we need to prove that if $ A \subset \emptyset$ it follows that $A = \emptyset$.

Proof
Suppose $A \neq \emptyset$ then we can find an $x \in A$ such that $x \notin \emptyset$. Since $A \subset \emptyset$ it follows that $x \notin A$ hence we have a contradiction. Therefore our assumption was false and thus $A = \emptyset$.

Yes, you are right.

Nice, thank you very much! (Smile)

 

[evinda]

Yes, you are right.

Could we also say it like that? (Thinking)

It is known that $\varnothing \subset B$, for all sets $B$.

Therefore, $\varnothing \subset A$.
We also now that $A \subset \varnothing$.

So, we conclude that $A=\varnothing$.

 

[Evgeny.Makarov]

We also now that $A \subset \varnothing$.

We now... what? The suspense is killing me! This reminds me of the "I Accidentally…" meme. (Smile)

If you mean that we assumed that $A\subset\varnothing$ (I prefer to use $\subseteq$ for possibly nonstrict inclusions), then you are correct. The only thing is you have to show that $A\subseteq B$ and $B\subseteq A$ imply $A=B$. This is obvious, but so is the original statement, so it's difficult to say whether its proof is required.

 

[evinda]

We now... what? The suspense is killing me! This reminds me of the "I Accidentally…" meme. (Smile)

(Giggle)

If you mean that we assumed that $A\subset\varnothing$ (I prefer to use $\subseteq$ for possibly nonstrict inclusions), then you are correct. The only thing is you have to show that $A\subseteq B$ and $B\subseteq A$ imply $A=B$. This is obvious, but so is the original statement, so it's difficult to say whether its proof is required.

Could prove it like that? (Thinking)

$A \subseteq \varnothing \leftrightarrow \forall x(x \in A \rightarrow x \in \varnothing)$

$\varnothing \subseteq A \leftrightarrow \forall x(x \in \varnothing \rightarrow x \in A)$

$A \subseteq \varnothing \wedge \varnothing \subseteq A \leftrightarrow \forall x(x \in A \leftrightarrow x \in \varnothing) \overset{\text{ axiom of extensionality }}{ \rightarrow } A=\varnothing$

 

[Evgeny.Makarov]

Yes, this is perfect.

 

[evinda]

Yes, this is perfect.

Nice! Thanks a lot! (Happy)