Thread: a little bit of functions on sets

1. a little bit of functions on sets

1. (a) For an arbitrary set Y , determine $Y^\emptyset$
(b) For an arbitrary set X, determine $\emptyset ^X$

(a) I couldn't figure it out
(b) I think it's the empty set but I can't show how
if anyone could show me how I would greatly appreciate it

2. Originally Posted by jbpellerin
1. (a) For an arbitrary set Y , determine $Y^\emptyset$
(b) For an arbitrary set X, determine $\emptyset ^X$

(a) I couldn't figure it out
(b) I think it's the empty set but I can't show how
if anyone could show me how I would greatly appreciate it
clarify what you mean. by the notation $B^A$, do you mean the set of functions from the set $A$ to the set $B$?

3. We can find disagreement on this question.
Paul Halmos claims the following.
“(i) $Y^\emptyset$ has exactly one element, namely $\emptyset$, whether $Y$ is empty or not and (ii) if $X \ne \emptyset$, then $\emptyset ^X = \emptyset$".

4. Originally Posted by Plato
We can find disagreement on this question.
Paul Halmos claims the following.
“(i) $Y^\emptyset$ has exactly one element, namely $\emptyset$, whether $Y$ is empty or not and (ii) if $X \ne \emptyset$, then $\emptyset ^X = \emptyset$".
is my interpretation of the problem correct? in other words, is Halmos saying the set of function from the empty set to another set is empty, and thus, represented by the empty set?

it seems strange that we should require X is nonempty in part (ii), since that is just a special case of part (i) (where Y is empty).

how exactly is this claim justified? what are the other interpretations?

5. Originally Posted by Jhevon
how exactly is this claim justified? what are the other interpretations?
All I can say is read his book: Naive Set Theory.
I do not defend his answer.

6. Originally Posted by Plato
All I can say is read his book: Naive Set Theory.
I do not defend his answer.
For me a function is a set of ordered pairs: $f:A \mapsto B \Rightarrow \quad f \subseteq A \times B$.
I am not sure what $\emptyset \times Y\quad \vee \quad Y \times \emptyset$ mean.