When is the set $B \times B$ a function?

[evinda]

Hi! (Smile)

Let $B$ be a nonempty set. Does it stand that $\bigcap \mathcal{P}B=\mathcal{P} \bigcap B$? Is the set $B \times B$ always a function? If not, what condition should $B$ satisfy, so that the relation $B \times B$ is a function?

Let $x \in \bigcap \mathcal{P}B$. Then $\forall b \in \mathcal{P}B(x \in b)$.
We want to show that $x \in \mathcal \bigcap B$, so, that $x \subset \bigcap B$.
How could we do this? (Thinking)

I thought that the set $B$ is not always a set, only when this condition is satified: if $<d,e> \in B$ and $<d,f> \in B$, then $e=f$. Am I right? (Thinking)

 

[Evgeny.Makarov]

Let $B$ be a nonempty set. Does it stand that $\bigcap \mathcal{P}B=\mathcal{P} \bigcap B$? Is the set $B \times B$ always a function? If not, what condition should $B$ satisfy, so that the relation $B \times B$ is a function?

It would be nice at least to start a new paragraph for the second question.

$\bigcap\mathcal{P}B=\varnothing$ because $\varnothing\in\mathcal{P}B$.

I thought that the set $B$ is not always a set

I think a set is always a set. (Smile) The set $B\times B$ is a function iff $|B|\le 1$.

only when this condition is satified: if $<d,e> \in B$ and $<d,f> \in B$, then $e=f$.

If $B$ is a relation, then you are right.

 

[evinda]

$\bigcap\mathcal{P}B=\varnothing$ because $\varnothing\in\mathcal{P}B$.

Could you explain me further why $\bigcap\mathcal{P}B=\varnothing$? (Worried)

I think a set is always a set. (Smile) The set $B\times B$ is a function iff $|B|\le 1$.

(Blush) How do we conclude this?

If $B$ is a relation, then you are right.

So, this condition doesn't stand just for sets, right? (Thinking)

 

[Evgeny.Makarov]

Could you explain me further why $\bigcap\mathcal{P}B=\varnothing$?

How do we conclude this?

You should consider examples to see why this is so.

So, this condition doesn't stand just for sets, right?

Could you say again what statement precisely are we considering?

 

[evinda]

You should consider examples to see why this is so.

I will think about it... (Thinking)

Could you say again what statement precisely are we considering?

We want a condition for $B$, so that $B \times B$ is a function.

I thought that $B \times B$ is a function if the following stands:

$$xBy \wedge xBz \rightarrow y=z$$

But, this stands only for relations, right?

How can we find then a condition?

 

[Evgeny.Makarov]

I thought that $B \times B$ is a function if the following stands:

$$xBy \wedge xBz \rightarrow y=z$$

This is not correct. What does $xBy$ even mean?

 

[evinda]

This is not correct. What does $xBy$ even mean?

I thought that it means $\langle x,y \rangle \in B$. Am I wrong? (Worried)

 

[Evgeny.Makarov]

I thought that it means $\langle x,y \rangle \in B$. Am I wrong?

OK, this is the standard definition. Though $xBy$ would not make sense if $B$ is a set of integers.

I thought that $B \times B$ is a function if the following stands:

$$xBy \wedge xBz \rightarrow y=z$$

Where did you get this from? What if $B=\{1,2\}$, then what does $xBy$ mean?

 

[evinda]

OK, this is the standard definition. Though $xBy$ would not make sense if $B$ is a set of integers.

Where did you get this from? What if $B=\{1,2\}$, then what does $xBy$ mean?

I don't know.. (Worried) So, it doesn't stand for any set, only for relations, right? (Thinking)

 

[Evgeny.Makarov]

So, it doesn't stand for any set, only for relations, right?

I am still not sure what exactly you mean by "it" in the quote above. You could get a better idea if you considered some examples.

 

[evinda]

I am still not sure what exactly you mean by "it" in the quote above. You could get a better idea if you considered some examples.

I meant the condition: $xRy \wedge xRz \rightarrow y=z$.

 

[Evgeny.Makarov]

I meant the condition: $xRy \wedge xRz \rightarrow y=z$.

No, you didn't. In post #9, you said, "it doesn't stand". To stand (the correct word is "to hold" or "be true"), you need to provide the exact value of $R$. Only when $R$ is given, you can say whether $xRy \wedge xRz \rightarrow y=z$ holds or nor. In post #5, rather, you said: "$B\times B$ is a function if the following stands: $xBy\land xBz\to y=z$. This statement (not "sentence") does not make sense when $B$ is not a set of pairs, e.g., when $B=\{1,2\}$. So I am still not sure about the exact statement you mean. You could get a better idea of it if you considered some examples.