Answer:
Elements are of the form
(i) ![[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}](https://tex.z-dn.net/?f=%5Ba%2Ca%5D%3D%5C%7B%5Bx%2Cy%5D%20%3A%20a%5Cleq%20x%5Cleq%20a%2C%20a%5Cleq%20y%5Cleq%20a%3B%20a%5Cin%20%5Cmathbb%20R%5C%7D)
(ii) 
(iii)![(a,a]=\{(x,y] :a](https://tex.z-dn.net/?f=%28a%2Ca%5D%3D%5C%7B%28x%2Cy%5D%20%3Aa%3Cx%5Ceq%20a%2Ca%3Cy%5Ceq%20a%3B%20%20a%5Cin%20%5Cmathbb%20R%5C%7D)
(iv)
(v) (a,b) where a>b=
(vi) [a,b] where a>b=![\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}](https://tex.z-dn.net/?f=%5C%7B%5Bx%2Cy%5D%20%3A%20a%5Cgeq%20x%5Cgeq%20b%2Ca%5Cgeq%20y%5Cgeq%20b%3Ba%3Eb%2Ca%2Cb%20%5Cin%20%5Cmathbb%20R%5C%7D)
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.
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,
![[a,b]=\{[5,0],[5,1],[5,2].....\}](https://tex.z-dn.net/?f=%5Ba%2Cb%5D%3D%5C%7B%5B5%2C0%5D%2C%5B5%2C1%5D%2C%5B5%2C2%5D.....%5C%7D)
e.t.c
So this set is connected and we know singletons are connected in . Hence given set is empty.
. Thus, given f(x), we have 