# Proofs involving sets - True and False?

Can someone please help me with these True and False questions? I've tried them myself, but I'm not very good at discrete math... Thank you in advance!

1. Any set $A$ and $B$ with $B\subseteq A$ and $f: B \to A$ be $1$-$1$ and onto, then $B = A$

False?

2. Let $A$ and $B$ be nonempty sets and $f:A \to B$ be a $1$-$1$ function. Then $f(X\cap Y) = f(X)\cap f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$

True?

3. Let $A$ and $B$ be nonempty sets and $f:A \to B$ be a function. Then if $f(X\cap Y) = f(X)\cap f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$, then $f$ must be $1$-$1$.

False?

4. There is no one-to-one correspondence between the set of all positive integers and the set of all odd positive integers because the second set is a proper subset of the first.

False?

5. if $(A \cup B\subset A \cup C)$ then $B\subset C$

False?

6. If $A$, $B$, and $C$ are three sets, then the only way that $A \cup C$ can equal $B \cup C$ is $A = B$.

False?

7. If the product $A \times B$ of two sets $A$ and $B$ is the empty set , then both $A$ and $B$ have to be empty set.

False?

1. False. Take $A$ the natural numbers, $B=A\setminus\{0\}$ the positive ones, and $f$ subtraction of$~1$. (Is true for any finite set $A$ however.)

2. True. The injective $f$ is a bijection to $f(A)$, and applying a bijection commutes with $\cap$ and $\cup$.

3. True. If $f$ were non-injective, two distinct elements of $A$ would have the same image; taking their singleton sets then contradicts the hypothesis.

4. False. See 1.

5. False, if $A\supseteq B,C$ then the hypothesis is trivial, but the conclusion is not.

6. False again, same counterexample but with $C$ in the role of "big brother" $A$.

7. False, one of them being empty suffices. This one is at the heart of possible confusion about $k\times l$ matrices with $k=0$ or $l=0$ (possibly both). You need to distinguish various types of empty matrices if you want to define multiplication correctly (a $n\times 0$ matrix multiplied by a $0\times m$ matrix gives a non-empty, though zero, $n\times m$ matrix), but you cannot make this distinction if you define matrices (as does Bourbaki) as a map on $[k]\times[l]$ assigning entries to positions, because $[k]\times[0]=\emptyset=[0]\times[l]$.

(4) There is no one-to-one correspondence between the set of all positive integers and the set of all odd positive integers because the second set is a proper subset of the first.

FALSE.

There is one : consider the one-to-one correspondance $n \rightarrow 2n+1$, where $n \in \mathbb Z^+$

(5) If $A \cup B \subseteq A \cup C$, then $B \subseteq C$.

FALSE

By a counterexample; consider $A = \{ 1, 2, 3 \}$, $B = \{ 3 \}$ and $C = \{ 4 \}$. We have that : $A \cup B = \{ 1, 2, 3 \} \subseteq A \cup C = \{ 1, 2, 3, 4 \}$, but $B \nsubseteq C$.

(6) If A, B, and C are three sets, then the only way that $A \cup C$ can equal $B \cup C$ is $A = B$.
You have to check that $A = B$ iff $A \cup C = B \cup C$; to do it, apply the property : $A = B$ iff $A \subseteq B$ and $B \subseteq A$ to check it.
For sure, it is true that if $A = B$, then $A \cup C \subseteq B \cup C$ and $B \cup C \subseteq A \cup C$. Thus, we conclude that : if $A = B$, then $A \cup C = B \cup C$.
What about the other "direction" ? We have already "solved" it in (4). Consider : $A = \{ 1, 2, 3 \}$, $B = \{ 1, 2 \}$ and $C = \{ 3, 4 \}$. We have that $A \cup C = \{ 1, 2, 3, 4 \} = B \cup C = \{ 1, 2, 3, 4 \}$, but $A \ne B$.
Thus, we conclude that it is not true that : if $A \cup C = B \cup C$, then $A=B$.