# Understanding a question from the topic FUNCTIONS.

I was trying to solve some problems about functions when I got stuck with this problem.

I am unable to understand this. First and foremost, $$A=[1,2]$$ $$B=[3,4]$$

This seems to be a little daunting. Because, in the first step, they have written that, $$f(A) = \{1\}$$

To be honest, I am seeing this for the first time (ie. Giving a set as the input and getting another set as the output). Maybe, they are trying to say that f(A) means f(a),where a is an element of A? Then why do they give its range in the form of a set?

You can see that in the second step, they have written that

$$f(A\cap\ B)$$ =$$f(\emptyset\ )$$= $$\emptyset\$$

How is this even possible? Empty set is not a member of the set of real numbers. And the last two steps also. There, you can see that they have taken the inverse of f(A). Is this possible? What should I infer from this?

Perhaps I have some serious misunderstandings related to this. Could you be kind enough to resolve my confusion? (I mean, I have looked for a similar question in some books. But, it was hopeless.)

In short, I would like to have a detailed explanation of this question and its steps. Thank you first of all for spending your valuable time to help me understand this question.

• You are correct. If $S$ is a subset of the domain of a function $f$, then people often write $f(S)$ to denote $\{f(s)\,|\,s\in S\}$. With that in mind, we do indeed have $f(\emptyset)=\emptyset$.
– lulu
Commented Sep 15, 2021 at 15:14
• @lulu. So, I can say that $$f(A)$$ = {1} actually means that the elements of A correspond to the element 1. Then, How can I write its INVERSE? (since the same element corresponds to all the elements of A, it is not one one ) and how come it is (0,infinity)? Commented Sep 15, 2021 at 16:10
• No. You can say that $f(A)=\{f(a)\,|\,a\in A\}=\{1\}$. Similarly, you can say that $f^{-1}(1)=\{x\in \mathbb R\,|\,f(x)=1\}=(0,\infty)$.
– lulu
Commented Sep 15, 2021 at 16:14
• @lulu one more doubt, please. I was wondering about the elements of $$f^{-1}(1)$$ . Won't they be (1,1),(1,1.5),(1,1000) etc?. Doesn't that make it a relation, rather than a function? Commented Sep 16, 2021 at 3:52
• I agree that $f^{-1}$ is not a function, at least not in the usual sense, because functions are required to take single values to single values. When one writes, e.g., $f^{-1}(1)$, we mean $\{x\in \mathbb R\,|\,f(x)=1\}$. Thus $f^{-1}$ of a single value might be an infinite set, as it is in this particular example.
– lulu
Commented Sep 16, 2021 at 11:58