All categories

3 years ago

Let A and B be bounded subsets of R. (a) Why does sup(AUB) exist? (b) Prove that sup(AUB) = max

Mathematics

1 answer:

makvit [3.9K]3 years ago

3 0

Answer:

a) sup(AUB) exist because A and B are bounded.

The definition of sup(A)={x∈A/y∈A,y≤x}

If x=sup(A),x∈A ⇒ x∈(A∪B)

If z=sup(B), z∈B ⇒ z∈(A∪B)

b)The value of sup(A∪B)=max(sup(B),sup(A))

proof

x∈A∪B⇒x∈A ∨ x∈B⇒x≤sup(A) ∨ x≤sup(B) then x≤max{sup(A),sup(B)}

sup(A)≤sup(A∪B) and

sup(B)≤sup(A∪B)

By definition of max:

max{sup(A),sup(B)}≤sup(A∪B).

You might be interested in

Which table has greater constant of proportionality?

marissa [1.9K]

The table with greater proportionality is table 3.

<h3>What is constant proportionality?</h3>

The constant of proportionality is the ratio between two directly proportional quantities.

As, the constant proportionality is

k= y/x

For table 1,

k = 1/1 = 2/2 =1

For table 2,

k = 6/1= 12/2 = 2

For table 3,

k = 5/1 = 10/2 = 5

For table 4,

k = 4/1 = 8/2 = 4

Hence, table 3 is with greater proportionality.

Learn more about this constant of proportionality here:

brainly.com/question/8598338

#SPJ1

8 0

1 year ago

Hello I need help with number 1-12 please and thank you!!!!!

kati45 [8]

Answer:

Please see above

Step-by-step explanation:

I have to right stuff here but the answer is above.

6 0

2 years ago

Type nae nae for 5 marks

baherus [9]

Answer:

nae nae

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

4 0

3 years ago

PLZ HELP ASAP SOLID GEOMETRY

eduard

It should be E) none of the above because:

height = 0.5
Radius = 1
diameter = 2

7 0

2 years ago

Read 2 more answers