Test: Real Analysis- 2 - Mathematics MCQ

Correct Answer :- c

Explanation : The set A below is the set of all limit points.

A = {0}∪{1/a ∣a∈N}

Since x∉A, there exists an ϵ>0 for which the 

I=(x−ϵ,x+ϵ) has no real number of the form 1/a where a∈N. 

Also, ϵ is such that 0∉I, and hence, x−ϵ>0. 

These are hold for the interval I′=(x−ϵ/2,x+ϵ/2), too. 

Our goal is to show that there are only finite number of elements both in S and I′, if any. 

To do so, assume y=1/m+1/n (n,m∈N) to be an element of both S and I′, and without loss of generality, 1/n ≤ 1/m. 

Since I has no real number of the form 1/a (a∈N), then:

S = {1/m, ∣m∈N}

For every δ > 0, (-δ, δ) contains infinite number of elements of set S. Hence, {0} is the set of all limit point of the set S.

Let n(M)=student who studying mathematics

n(C)=student who studying chemistry

n(P)=student who studying Physics

∴n(M)=23,n(P)=24,n(C)=19,n(M∩P)=12,n(M∩C)=9,n(P∩C)=7,n(M∩P∩C)=4

Number of student who studying mathematics but not physic and chemistry

⇒n(M)−[(n(M∩C)+n(M∩P)]+n(M∩P∩C)

⇒23−[9+12]+4

⇒23−21+4=6

Number of student who studying chemistry but not physic and matehematics

⇒n(C)−[(n(M∩C)+n(P∩C)]+n(M∩P∩C)

⇒19−[9+7]+4

⇒19−16+4=7

Number of student who studying physics but not mathematics and chemistry

⇒n(P)−[(n(M∩P)+n(P∩C)]+n(M∩P∩C)

⇒24−[12+7]+4

⇒24−19+4=9

∴no. of student studying exactly one subject = 6+7+9 = 22.

Let A = { 1 ,2 ,3 ,4 } and R be a relation on A given by R = {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (3,1), (1,3)}. 
Since, (1, 1), (2, 2), (3, 3), (4, 4) ε R therefore R is reflexive.
Again since (1, 2), (3, 1) ε R and also (2,1), (1, 3) e R. Hence, R is symmetric. But clearly R clearly R is not transitive.

We are given that
X = { 1 ,2 ,3 ,4 ,5 } 
and Y= { 1 ,3 ,5 ,7 ,9 }
Since, (7, 9) ε R₄ but (7, 9) ∉ X x Y therefore R₄ is not a relation from X to Y.

Let us suppose that the set of real num ber R is countable, then R = {x₁, x₂, ..., x_n, ...}. Enclose each member x_n of R in an open interval I_n =
 of length where the sum of lengths of I_n's is 
But 
Implies, The whole real line is contained in the union of intervals whose lengths add up to 1. Which is a contradiction. Hence, R is uncountable.
Arrange the set of rationals according to increasing denominators as
Then the one-one correspondence can be indicated as
which shows that the set of rational number in [0, 1] is countable. Now, the set of all rational numbers is the union , where A_i is the set of rationals which can be written with denominator i. Then is A, = 
Each A_i' is equivalent to the set of all positive integers and hence countable.

The set l has no limit point, for a nbd of m ε l, contains no point of l other than m. Thus the derived set of l is the null set φ. The set N has no limit point for a nbd  of m ε N, contains no points of N other than m. Thus, the derived set of N is null set φ. Every point of the set Q of rationals is a limit point, for between any two rationals there exist an infinity of rational.
Further every point of R is a limit point, for every nbd of any of its points contains an infinite member of R.

Let G₁, and G₂ be two open sets. Then, if G₁ ∩ G₂ = φ, it is open
if G₁ ∩ G₂ ≠ φ, let x ε G₁ ∩ G₂
implies x ε G₁, and x ε G₂
implies G₁, G₂ are nbd of x.
implies G₁ ∩ G₂ is a nbd of x.
but since x is any point o f G₁ ∩ G₂ therefore G₁ ∩ G2 is a nbd o f each of its points. Hence G₁ ∩ G2 is open.
Now, consider the open sets

implies which is not an Open set.
implies The intersection of arbitrary number of open set need not be open. Next, let G be the union of an arbitrary family of open sets, A being an index set. To prove that G is an open set, we shall show that for any point x ε G, it contains an open interval containing x.
Let x ε G impliesatleast one member, say of F such that x ε . Since, is an set, an open interval ln such that
Thus, the set G contains an open interval containing any point x of G. Hence, G is an open set.

Let (a, b) be an interval and x ε (a, b) which implies that a < x < b.

Let c, d be two numbers sueh that a < c < x and n < d < h
implies a < c < x < d < b
implies x ε (c, d) ⊂ (a, b)
Thus, the given interval (a, b) contains an open interval containing the point x and is therefore a nbd of x. Hence, the open interval is a nbd ol each o f its points and is therefore an open set. Hence, every point of an open interval is an interior point. Now ‘o’ is an interior point o f the set

which does not lie in this
set. Hence, the set

is not open.

Since, for the set N, I and Z there is no open interval which is contained in N or I or Z as every open interval contains rational and irrational number but the set of interior points of R is R.

A set can be a nbd of a point if it contains an open interval containing the point. Since, an interval necessarily contains an infinite number of points. Therefore in order that a set be a nbd of a point it necessarily contain an infinity of points. Thus, a finite set cannot be a nbd of any points. Hence, no points of finite non-empty set is an interior point.

If a set contains ‘n’ elements, then the number of proper subsets of the set is 2ⁿ - 1.
If A = {p, q} the proper subsets of A are [{ }, {p}, {q}]
⇒ Number of proper subsets of A are 3 = 2² - 1 = 4 - 1
In general, number of proper subsets of a given set = 2^m - 1, where m is the number of elements.