TIFR CSE 2017 | Part A | Question: 10

Question

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

• A. $f$ cannot be one-to-one (injective)
• B. $f$ cannot be onto (surjective)
• C. $f$ is both one-to-one and onto (bijective)
• D. there is no such $f$ possible
• E. if such a function $f$ exists, then $A$ is infinite

Answer 1

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$(B)$

Comments

Thanks @vizzard110

I got it now. :)

So we just have a function that has domain as the elements of A and co-domain as the subsets of A. (Mapping can be anything)

Since |P(A)|>|A| we cannot make the function onto.

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@rohith1001 you can take any possible mapping of your own choice but you can always see that onto is not possible.

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how the mapping is exactly done here?  element $1,2$ can be mapped with $\phi$ and we say it is not one one function??

Answer 2

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)