# More sets and functions

• Aug 21st 2007, 12:36 PM
shilz222
More sets and functions
(1) Let $f: X \to Y$ be a function and $A_{1}, A_{2} \in \mathcal{P}(X)$. (i) Prove that $A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) \subseteq \overrightarrow{f}(A_{2})$. Prove that the converse is not universally true. Give a simple condition on $f$ which is equivalent to the converse. (ii) Prove that $\overrightarrow{f}(A_{1} \cap A_{2}) \subseteq \overrightarrow{f}(A_{1}) \cap \overrightarrow{f}(A_{2})$. Prove that equality is not universally true. (iii) Prove that $\overrightarrow{f}(A_{1} \cup A_{2}) = \overrightarrow{f}(A_{1}) \cup \overrightarrow{f}(A_{2})$.

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

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

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

Thanks
• Aug 21st 2007, 12:42 PM
Plato
(i) Prove that $A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) = \overrightarrow{f}(A_{2})$.
That is not true!

This is true.
(i) Prove that $A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) \subseteq \overrightarrow{f}(A_{2})$.
• Aug 21st 2007, 12:44 PM
shilz222
I meant to say that $\overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$.
• Aug 21st 2007, 01:17 PM
Plato
Quote:

Originally Posted by shilz222
I meant to say that $\overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$.

Here are some notes.
• Aug 23rd 2007, 12:45 PM
shilz222
For 1(i) the converse $\overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2)$ is not universally true.

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

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

Is this correct?
• Aug 23rd 2007, 01:29 PM
Plato
Quote:

Originally Posted by shilz222
Define a function $f(A) = \emptyset$.

That is not allowed.
$f(A) = \emptyset$ makes no sense!
Functions are sets of ordered pairs, a subset of some $A \times B,\;A \not= \emptyset \wedge B \not= \emptyset$.
• Aug 23rd 2007, 01:36 PM
shilz222
Ok if $A_ 1 = \{ a,b,c,d \} \ \text{and} \ A_2 = \{c,d,e,f,g,l,m \}$ and define a function $f(c) = 6$. Then $\overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2) = 6$ but $A_1 \not \subseteq A_2$.
• Aug 23rd 2007, 01:44 PM
Plato
Suppose that $A = \{ 1,2,3,4\} ,\;B = \{ h,j,k,m\}$ then define a function $f:A \mapsto B$ by $f = \left\{ {\left( {1,j} \right),\left( {2,m} \right),\left( {3,k} \right),\left( {4,j} \right)} \right\}$.
Let $A_1 = \left\{ {1,3} \right\}\;\& \;A_2 = \left\{ {2,3,4} \right\}$ note that $\overrightarrow f \left( {A_1 } \right) \subseteq \overrightarrow f \left( {A_2 } \right)\quad but\quad A_1 \not\subseteq A_2$
• Aug 23rd 2007, 02:12 PM
shilz222
Was my example correct?
• Aug 23rd 2007, 02:30 PM
Plato
Quote:

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.
• Aug 23rd 2007, 02:40 PM
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?
• Aug 23rd 2007, 02:55 PM
Plato
Quote:

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.

Quote:

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.
• Aug 23rd 2007, 02:56 PM
topsquark
Quote:

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
• Aug 23rd 2007, 08:06 PM
ThePerfectHacker
Quote:

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.
• Aug 24th 2007, 11:08 AM
shilz222
For 1(i) the condition on $f$ is that it has to be injective?