Direct Image Of Union Of Sets Proof: F(A U B) = F(A) U F(B) - YouTube

+-- | a | b | c | result | |.Even with ConstraintKinds, I think the barrier is going to be quantifying over the "type function" from the arguments to the results. What you want is for f to map a -> b and c -> d, and to take a -> b -> (c, d), but I don't think there's any way to quantify over that relationship with full generality.Similar questions to practice: function-85751.html functions-problem-need-help-93184.html let-the-function-g-a-b-f-a-f-b-143311.html. Location: Canada. Re: For which of the following functions is f(a+b)=f(b)+f(a) [#permalink] 06 Feb 2018, 14:50.

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The coefficients a, b, c, e.t.c. are known constant numbers that determine the shape of the polynomial. x is a variable for which we can choose values. The highest power of x that appears in the polynomial is called the "degree" of the polynomial. We write the expression with the highest power of x first then...Пример 1. Для формулы A/\ (B \/ ¬B /\¬C) постройте таблицу истинности.Let there be a function $f:R→R$. Let $A$ & $B$ be two subsets of $R$. I don't know how to prove it. Also, are there certain conditions that the function $f$ or the sets $A$ & $B$ should follow for the above condition to be true?

For which of the following functions is f(a + b) = f(a) + f(b) for all | Forum

Set Theory Proof A is a subset of B if and only if A intersect B = A.

functions - Proof of $f(A∩B)⊆f(A)∩f(B)$ - Mathematics Stack Exchange Stack Exchange Network

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Let there be a function $f:R→R$. Let $A$ & $B$ be two subsets of $R$. From my experience, I know that $f(A∩B)⊆f(A)∩f(B)$.

I don't know how to prove it. Also, are there certain conditions that the function $f$ or the sets $A$ & $B$ should follow for the above condition to be true?

egreg

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asked Apr 16 '17 at 14:05

Salman QureshiSalman Qureshi

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Take $y\in f(A\cap B)$. Then, by definition, there exists $x\in A\cap B$ such that $y=f(x)$;

since $x\in A$, we have $y=f(x)\in f(A)$, since $x\in B$, we have $y=f(x)\in f(B)$.

Therefore $y=f(x)\in f(A)\cap f(B)$

Equality does not necessarily hold. Take $f\colon\{1,2\}\to\{0\}$ (the only possible map); take $A=\{1\}$ and $B=\{2\}$. Then $f(A\cap B)=f(\emptyset)=\emptyset$, whereas $f(A)\cap f(B)=\{0\}$.

answered Apr 16 '17 at 14:14

egregegreg

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Definitions:

$f(A \cap B) = \{f(x): x \in A \cap B\}$ $f(A) \cap f(B) = \{f(x): x \in A\} \cap \{f(x): x \in B\}$

Now, $A \cap B \subseteq A$ and $A \cap B \subseteq B$.

Therefore, $f(A \cap B) \subseteq f(A)$ and $f(A \cap B) \subseteq f(B)$.

And the result follows from messing around with the logical definitions of $\subseteq$ and $\cap$.

answered Apr 16 '17 at 14:06

DHMODHMO

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A proof by elements is the safe way:

Let $y\in f(A\cap B)$. By definition, $y = f(x)$ for some $x\in A\cap B$. Therefore $f(x)\in A$ and $f(x)\in B$, which means $y=f(x)\in f(A)\cap f(B)$.

answered Apr 16 '17 at 14:11

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Let $y \in f(A \cap B)$. Then $y=f(x)$ for some $x \in A \cap B$.

Since $x \in A$ and $y=f(x)$ we get $y \in f(A)$.

Since $x \in B$ and $y=f(x)$ we get $y \in f(B)$.

Therefore, $y \in f(A) \cap f(B)$.

This shows $f(A\cap B) \subseteq f(A) \cap f(B)$.

answered Apr 16 '17 at 14:11

N. S.N. S.

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Directly by definition you can prove it.

Let $y \in f(A \cap B)$. (This is because $X \subset Y$ means every element of $X$ is of $Y$; so to prove $X \subset Y$ we take an arbitrary one out of $X$ to see if that one is in $Y$.) Then by definition there is some $x \in A \cap B$ ensuring $y = f(x)$. Then note that $x \in A \cap B$ implies $x \in A$ and $x \in B$, implying that $f(x) \in f(A)$ and $\in f(B)$. So $y \in f(A) \cap f(B)$.

answered Apr 16 '17 at 14:12

MegadethMegadeth

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