Answer:
Elements are of the form
(i)
(ii) ![[a,b)=\{[x,y) :a\leq x](https://tex.z-dn.net/?f=%5Ba%2Cb%29%3D%5C%7B%5Bx%2Cy%29%20%3Aa%5Cleq%20x%20%3Ca%2Ca%5Cleq%20y%20%3Ca%3B%20a%5Cin%20%5Cmathbb%20R%5C%7D)
(iii)
(iv)
(v) (a,b) where a>b=
(vi) [a,b] where a>b=
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.
![\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}](https://tex.z-dn.net/?f=%5C%7B%5Ba%2Ca%5D%20%3A%20a%5Cin%20%5Cmathbb%20R%5C%7D%3D%5C%7B%5B0%2C0%5D%2C%5B1%2C%201%5D%2C%5B-1%2C-1%5D%2C%5B2%2C2%5D%2C%5B-2%2C-2%5D%2C%5B3%2C3%5D%2C%5B-3%2C-3%5D%2C........%5C%7D)
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if
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.
![[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........](https://tex.z-dn.net/?f=%5B%201%2C%201%29%2C%5B-1%2C-1%29%2C%5B2%2C2%29%2C%5B-2%2C-2%29%2C%5B3%2C3%29%2C%5B-3%2C-3%29%2C........)
Since there is a singleton element a of real numbers withis the set, this set is empty.
Because there is no increment so if
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.
![( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........](https://tex.z-dn.net/?f=%28%201%2C%201%5D%2C%28-1%2C-1%5D%2C%282%2C2%5D%2C%28-2%2C-2%5D%2C%283%2C3%5D%2C%28-3%2C-3%5D%2C........)
Since there is a singleton element a of real numbers, this set is empty.
Because there is no increment so if
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.
![( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........](https://tex.z-dn.net/?f=%28%201%2C%201%29%2C%28-1%2C-1%5D%2C%282%2C2%5D%2C%28-2%2C-2%5D%2C%283%2C3%5D%2C%28-3%2C-3%5D%2C........)
Because there is no increment so if
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,
e.t.c
So this set is connected and we know singletons are connected in
. Hence given set is empty.
(vi) [a,b] where
.
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,
e.t.c
So this set is connected and we know singletons are connected in
. Hence given set is empty.