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+3 votes

For a set $A$ define $\mathcal{P}(A)$ to be the set of all subsets of $A$. For example, if $A = \{1, 2\}$ then $\mathcal{P} (A) = \{ \emptyset, \{1, 2\}, \{1\}, \{ 2 \} \}$. Let $A \rightarrow \mathcal{P}(A)$ be a function and $A$ is not empty. Which of the following must be TRUE?

  1. $f$ cannot be one-to-one (injective)
  2. $f$ cannot be onto (surjective)
  3. $f$ is both one-to-one and onto (bijective)
  4. there is no such $f$ possible
  5. if such a function $f$ exists, then $A$ is infinite
asked in Set Theory & Algebra by Veteran (92.7k points) 987 2344 3118 | 178 views

2 Answers

+10 votes
Best answer

Even if it can be one-to-one in the following way,

But, It cannot be onto,because, the number of elements in domain $(A)$  $<$ the number of elements in co-domain ($P(A)$) . For a function to be onto, the domain should be able to cover all elements of co-domain with each element of the domain having exactly one image in co-domain.
so option(B)

answered by Veteran (12.5k points) 12 52 155
edited by

Nice solution.Thanks :)

a function output for a single value is a single value only not a set of values So I think it is wrong
0 votes

If the number of elements in Co-domain is greater than number of elements of domain, then the function cannot be onto...

If the number of elements in domain is greater than codomain,then the function cannot be one-to-one.

Here number of elements in co-domain is greater than domain,so obviously there will be atleast one element in co-domain which will not have a preimage in the domain. But it can be one-to-one... 

So Option B)


answered by Loyal (4.9k points) 3 35 121

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