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The complete statement is: Let $(A,≤)$ be a well-ordered nonempty set and let $f$ be the function of $P(A)-${$Ø$} in $A$ defined for: $f(S) =$ first element of $S$.

(a) Prove that $f$ is surjective.

(b) Prove that if $| A | > 1$ then $f$ is not injective.

Clarification: Let $A$ be a set, $P(A)$ denotes the set of parts of $A$. The elements of $P (A)$ are the subsets of $A$:

$P(A)=$ {$x:x$ is a subset of $A$}

To understand the problem,I have create a small example that is the following:

$A=${$1,2,3$} then $P(A)=${$Ø$,{$1$},{$2$},{$3$},{$1,2$},{$1,3$},{$2,3$},{$1,2,3$},}

Then, $f$({$1$})$=$$f$({$1,2$})$=$$f$({$1,3$})$=$$f$({$1,2,3$})$=1$

$f$({$2$})$=$$f$({$2,3$})$=2$

$f$({$3$})$=3$

So for this small example it is verified that every element of $A$ is an image of some element of $P(a)$ then, $f$ is surjective

But I don't know how to demonstrate the exercise in a general way so that's my question.

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    $\begingroup$ For your first question, your example should help you understand how to find an antecedent of any element $a\in A$. The simplest subset containing $a$ as its first element is..? For the second question, consider two distinct elements $a$ and $a'$, and assume that $a<a'$. Can you build two distinct subsets, using $a$ and $a'$, which have the same image by $P$? $\endgroup$ – Suzet Jun 21 '18 at 3:52
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    $\begingroup$ The simplest subset that contains containing $a$ as its first element is is the subset that contains only that element, {$a$} $\endgroup$ – Ayesca Jun 21 '18 at 4:04
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For $x\in A$, we have $f(\{x\})=x$.

It follows that $f$ is surjective.

Next, suppose $|A| > 1$.

Let $x,y\in A$, with $x\ne y$.

Without loss of generality, assume $x < y$.

Then $f(\{x,y\}) = x$, and $f(\{x\}) = x$, so $f$ is not injective.

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