Question

Show that the set of all irrational numbers is not of measure zero

Answer 1

By definition of measure zero, u must show that there exists ϵ>0ϵ>0 such that the set AA of irrationals in [0,1][0,1] cannot be covered by a countable collection of intervals [a1,b1],[a2,b2],…[a1,b1],[a2,b2],… with total length less than ϵϵ.

So I take ϵ=1/2ϵ=1/2, and suppose for contradiction that the countable collection of intervals [a1,b1],[a2,b2],…[a1,b1],[a2,b2],… covers AA and (b1−a1)+(b2−a2)+…<1/2(b1−a1)+(b2−a2)+…<1/2.

Actually I haven't learned any measure theory, only the definition of measure zero in real analysis. Is it possible to proceed using only that? (By the way, sure, I can show Q∩[0,1]Q∩[0,1] has measure zero

Answer 2

Answer: The proof for the above given question is shown below

Step-by-step explanation: Let A = [ 0 , 1 ]

I = set of irrational numbers in IR

Q = set of rational numbers in IR

then, A ∩ I ⊆ I

Hence, μ ( A ∩ I ) ≤ μ ( I ) ......... 1

Now μ ( A ) = length of interval [ 0 , 1 ]

= μ( [ 0 , 1 ] )

= 1 - 0

= 1

Now, A = ( A ∩ I ) ∪ ( A ∩ Q )

A ∩ I and A ∩ Q both are adjoint sets

A & Q both are measurable and

A - ( A ∩ Q ) = A ∩ I is also measurable

Therefore A ∩ Q and A ∩ I both are disjoint measurable sets

Hence

μ ( A ) = μ [ ( A ∩ I ) ∪ ( A ∩ Q ) ]

            = μ ( A ∩ I ) + μ ( A ∩ Q )

μ ( A ) - μ ( A ∩ Q ) = μ ( A ∩ I )

1 - 0 = μ ( A ∩ I )

Since μ ( Q ) = 0 and A ∩ Q ⊆ Q

μ ( A ∩ I )  = 1

From equation 1 we have

μ ( I ) ≥ μ ( A ∩ I )

μ ( I ) ≥ 1

Hence set of irrational numbers is not of measure zero.

The below given two links shows what is an irrational number and set of irrational numbers.

https://brainly.in/question/54092525

https://brainly.in/question/14909551

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