# Thread: More sets and functions

1. ## More sets and functions

(1) Let $\displaystyle f: X \to Y$ be a function and $\displaystyle A_{1}, A_{2} \in \mathcal{P}(X)$. (i) Prove that $\displaystyle A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) \subseteq \overrightarrow{f}(A_{2})$. Prove that the converse is not universally true. Give a simple condition on $\displaystyle f$ which is equivalent to the converse. (ii) Prove that $\displaystyle \overrightarrow{f}(A_{1} \cap A_{2}) \subseteq \overrightarrow{f}(A_{1}) \cap \overrightarrow{f}(A_{2})$. Prove that equality is not universally true. (iii) Prove that $\displaystyle \overrightarrow{f}(A_{1} \cup A_{2}) = \overrightarrow{f}(A_{1}) \cup \overrightarrow{f}(A_{2})$.

(i) So $\displaystyle x \in A_1 \Rightarrow x \in A_2$. Then $\displaystyle \overrightarrow{f}(A_{1}) = \{f(x) | x \in A_1 \} \Rightarrow \overrightarrow{f}(A_{2}) = \{f(x) | x \in A_2 \}$. Thus $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$. The converse is $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2) \Rightarrow A_1 \subseteq A_2$. Suppose that $\displaystyle A_{1} = \{c \}$ and $\displaystyle A_{2} = \{d \}$. Let $\displaystyle f(a) = a$. Then $\displaystyle \overrightarrow{f}(A_1) = \overrightarrow{f}(A_2) = \emptyset$ but $\displaystyle A_1 \not \subseteq A_2$. Is this correct? What would be the condition on $\displaystyle f$ which is equivalent to its converse?

(ii) $\displaystyle \overrightarrow{f}(A_{1} \cap A_{2}) = \{ f(x) | x \in A_{1} \cap A_2 \}$ which is a subset of $\displaystyle \overrightarrow{f}(A_{1}) \cap \overrightarrow{f}(A_{2})$. What would be a counterexample to show inequality?

(iii) This is basically the same as part (ii) except with an $\displaystyle \text{or}$?

Thanks

2. (i) Prove that $\displaystyle A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) = \overrightarrow{f}(A_{2})$.
That is not true!

This is true.
(i) Prove that $\displaystyle A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) \subseteq \overrightarrow{f}(A_{2})$.

3. I meant to say that $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$.

4. Originally Posted by shilz222
I meant to say that $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$.
Here are some notes.

5. For 1(i) the converse $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$ is not universally true.

Suppose that $\displaystyle A_{1} = \{a, b \}$ and $\displaystyle A_2 = \{c,d \}$. Define a function $\displaystyle f(A) = \emptyset$. Then $\displaystyle \overrightarrow{f}(A_1) = \overrightarrow{f}(A_2) = \emptyset$ but $\displaystyle A_{1} \not \subseteq A_{2}$.

A condition on $\displaystyle f$ which is equivalent to its converse is that $\displaystyle f(A) \neq \emptyset$.

Is this correct?

6. Originally Posted by shilz222
Define a function $\displaystyle f(A) = \emptyset$.
That is not allowed.
$\displaystyle f(A) = \emptyset$ makes no sense!
Functions are sets of ordered pairs, a subset of some $\displaystyle A \times B,\;A \not= \emptyset \wedge B \not= \emptyset$.

7. Ok if $\displaystyle A_ 1 = \{ a,b,c,d \} \ \text{and} \ A_2 = \{c,d,e,f,g,l,m \}$ and define a function $\displaystyle f(c) = 6$. Then $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2) = 6$ but $\displaystyle A_1 \not \subseteq A_2$.

8. Suppose that $\displaystyle A = \{ 1,2,3,4\} ,\;B = \{ h,j,k,m\}$ then define a function $\displaystyle f:A \mapsto B$ by $\displaystyle f = \left\{ {\left( {1,j} \right),\left( {2,m} \right),\left( {3,k} \right),\left( {4,j} \right)} \right\}$.
Let $\displaystyle A_1 = \left\{ {1,3} \right\}\;\& \;A_2 = \left\{ {2,3,4} \right\}$ note that $\displaystyle \overrightarrow f \left( {A_1 } \right) \subseteq \overrightarrow f \left( {A_2 } \right)\quad but\quad A_1 \not\subseteq A_2$

9. Was my example correct?

10. Originally Posted by shilz222
Was my example correct?
Do you see how different in detail my example is from the one you tried to give? To be honest, I would not accept it. It lacks detail that shows that you really understand how functions operate. But I am not grading you. So I cannot say if it correct or not; just I do not find it so.

11. How do you become better at these types of problems and math in general? Do you just need to read a lot of books an do a lot of problems and ask interesting questions? I mean is math meant to be learned slowly?

12. Originally Posted by shilz222
How do you become better at these types of problems and math in general?
I have always advised students to simply copy out proofs from well-written textbooks, a book at the correct level. Set theory & logic is the basis of all proofs.

Originally Posted by shilz222
I mean is math meant to be learned slowly?
Well it certainly is not meant to be approached in a shotgun way that your assortment of problems indicates you are trying.

13. Originally Posted by shilz222
How do you become better at these types of problems and math in general? Do you just need to read a lot of books an do a lot of problems and ask interesting questions? I mean is math meant to be learned slowly?
My own take on this is that Math is learned in basically the same manner as practically any other field. You need to work problems, ask questions, and work more problems. Some time for the information to "gel" helps, as well as seeing how the material applies to either real-life problems or how it is applied to other fields.

Just keep at it and be patient. You'll get it.

-Dan

14. Originally Posted by shilz222
I mean is math meant to be learned slowly?
Maybe the problem is that this is very abstract to thee. The first book on math having serious proofs I ever read was on number theory which is much less abstract. And perhaps this will make you understand what proofs are about. All set theory is is doing it much much more formally.

15. For 1(i) the condition on $\displaystyle f$ is that it has to be injective?