Proof or disproof: 1. Let A belongs to R and suppose A has neither an upper boun

Question

Proof or disproof:

1. Let A belongs to R and suppose A has neither an upper bound nor lower bound, then A=R

2. Suppose A belongs to R and A is not empty set, a also has a lower bound x. Then A has a largest lower bound b.

3. The Infimum of set P(all positive real numbers) is zero

4. the set R itself is a stepping set and so is P (all positive real numbers).

5. suppose A is a subset of B and B is a subset of R. If sup(B) exists, and A is not empty set, then sup(A) exists also, and sup(A) <=sup(B).

6. Let a,b belongs to R, and a<b. let set M={x belongs to R| a<x and x<b}. proof that supM =b, and infM=a.

Explanation / Answer

1) not true

let I(k) = (k, k+1) interval where k are integers then

take A = union of all I(k) for all k

this set neither has a upper bound nor a lower bound and A= R- Z   where Z is set of all integers

so A is not equal to R

2)

A has a lower bound a so

d(A, a)>=0

if it is 0 then a is the largestlower bound , b=a

if d(A,a) >0 then we can find x in R such that   0<d(A,x) < d(A,a)

and we keep doing this till we get d(A,x) = 0 then that x is our b

3)

let infeamum is x>0 , x in P

then x/2 is also in P and x/2<x so x is not infemum there is no such x

4)not true

5)

B<=sup(B)

A is subset of B so A<= sup(B)

so A is bounded and since A<=sup(B)
so sup(A)<=sup(B)

6)

M=(a,b)

so for x in M

x>a , and x<b

so supM =b, and infM=a

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