Let B be the set of rational numbers in the interval 0 1 and

Let B be the set of rational numbers in the interval [0, 1], and let {I_k}^n_k=1 be a finite collection of open intervals that covers B. Prove that Sigma^n_k=1 m*(I_k) greaterthanorequalto 1.

Solution

Proof :

Since 0 A, there is an open interval J₁ in {I_n} such that 0 J₁.

Let J₁ = (a₁, b₁). Note that a₁ < 0 and b₁ > 0.

If b₁ 1, then L(I) l( J₁) = b₁ a₁ 1. Suppose not. If a1₁ is rational, then I can find an open interval J₂ {I_n}   such that a₁ J₂.

If a₁ is irrational, I consider the following cases.

Case 1. There is an open interval J₂ such that a₁ J₂.

Case 2. There is an open interval J₂ such that a₁ is the right endpoint of J₂.

Case 3. Otherwise. We claim that Case 3 is impossible. Consider the following sub collection K₁, ..., Km where Ki {I In|a₁ < x for all x I}. Thus we select a open interval K₀ = (x, y) nearest a₁. Hence (a₁, x) Q cannot be covered by any elements of {I_n}, a contradiction. Hence we can find J₂ in Case 1 and Case 2.

Continue the process to find out J₃, .... Since {I_n}   is a finite covering of A, this process can be done in finite steps.

Hence L(I_n) J_n = (a_n+1 a_n) = a_m – a₁.

Note that a_m 1 and a_n 0.

Hence L(I_n)>1 , here we can replace L(I_n)=m(I_k)

So m(I_k)>1.

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