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

AI Thread Summary
The discussion centers on the conditions under which the Cartesian product $B \times B$ can be considered a function. It is established that $B \times B$ is a function if and only if the set $B$ contains at most one element, meaning that for any two pairs in $B$, the second elements must be equal. Participants also explore the relationship between the intersection of power sets and the empty set, concluding that $\bigcap \mathcal{P}B$ is indeed empty. Clarification is sought regarding the definitions and implications of certain statements, particularly in relation to sets versus relations. Overall, the conversation emphasizes the need for precise definitions when discussing mathematical concepts.
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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