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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?

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2 Answers 2

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  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 $[k]\times[0]=\emptyset=[0]\times[l]$.

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About :

(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 $n \in \mathbb Z^+$

About :

(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 $B \nsubseteq C$.

About :

(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 $A=B$.

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