Answer :
Answer:
Elements are of the form
(i) [tex][a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}[/tex]
(ii) [tex][a,b)=\{[x,y) :a\leq x <a,a\leq y <a; a\in \mathbb R\}[/tex]
(iii)[tex](a,a]=\{(x,y] :a<x\eq a,a<y\eq a; a\in \mathbb R\}[/tex]
(iv)[tex](a,a)=\{(x,y): a<x<b,a<x<b; a\in \mathbb R\}[/tex]
(v) (a,b) where a>b=[tex]\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}[/tex]
(vi) [a,b] where a>b=[tex]\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}[/tex]
Step-by-step explanation:
Given intervals are,
(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.
To show all its elements,
(i) [a,a]
Imply the set including aa from left as well as right side.
Its elements are of the form.
[tex]\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}[/tex]
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if [tex] a\in \mathbb R[/tex] then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].
(ii) [a,a)
This means given interval containing a by left and exclude a by right.
Its elements are of the form.
[tex][ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........[/tex]
Since there is a singleton element a of real numbers withis the set, this set is empty.
Because there is no increment so if [tex] a\in \mathbb R[/tex] then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).
(iii) (a,a]
It means the interval not taking a by left and include a by right.
Its elements are of the form.
[tex]( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........[/tex]
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if [tex] a\in \mathbb R[/tex] then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].
(iv) (a,a)
Means given set excluding a by left as well as right.
Since there is a singleton element a of real numbers, this set is empty.
Its elements are of the form.
[tex]( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........[/tex]
Because there is no increment so if [tex] a\in \mathbb R[/tex] then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).
(v) (a,b) where a>b.
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
[tex](a,b)=\{(5,0),(5,1),(5,2).....\}[/tex] e.t.c
So this set is connected and we know singletons are connected in [tex]\mathbb R[/tex]. Hence given set is empty.
(vi) [a,b] where [tex]a\leq b[/tex].
Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.
That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,
[tex][a,b]=\{[5,0],[5,1],[5,2].....\}[/tex] e.t.c
So this set is connected and we know singletons are connected in [tex]\mathbb R[/tex]. Hence given set is empty.