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!
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Any set $A$ and $B$ with $B\subseteq A$ and $f: B \to A$ be $1$-$1$ and onto, then $B = A$
False?
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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?
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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?
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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?
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if $(A \cup B\subset A \cup C)$ then $B\subset C$
False?
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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?
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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?
Best Answer
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.)
True. The injective $f$ is a bijection to $f(A)$, and applying a bijection commutes with $\cap$ and $\cup$.
True. If $f$ were non-injective, two distinct elements of $A$ would have the same image; taking their singleton sets then contradicts the hypothesis.
False. See 1.
False, if $A\supseteq B,C$ then the hypothesis is trivial, but the conclusion is not.
False again, same counterexample but with $C$ in the role of "big brother" $A$.
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 $[k]\times[0]=\emptyset=[0]\times[l]$.