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

evinda
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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)
 
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evinda said:
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 $\varnothing\in\mathcal{P}B$.

evinda said:
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 $|B|\le 1$.

evinda said:
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.
 
Evgeny.Makarov said:
$\bigcap\mathcal{P}B=\varnothing$ because $\varnothing\in\mathcal{P}B$.

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

Evgeny.Makarov said:
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? :confused:

Evgeny.Makarov said:
If $B$ is a relation, then you are right.

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

evinda said:
So, this condition doesn't stand just for sets, right?
Could you say again what statement precisely are we considering?
 
Evgeny.Makarov said:
You should consider examples to see why this is so.

I will think about it... (Thinking)

Evgeny.Makarov said:
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? :confused:
 
evinda said:
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?
 
Evgeny.Makarov said:
This is not correct. What does $xBy$ even mean?

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

evinda said:
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?
 
Evgeny.Makarov said:
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)
 
  • #10
evinda said:
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.
 
  • #11
Evgeny.Makarov said:
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: $xRy \wedge xRz \rightarrow y=z$.
 
  • #12
evinda said:
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.
 

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