prove a stronger version of proposition 2.3.7 suppose xn ia a sequence such that every subsequence xni has a subsequence xnmi that converges to x

Answer and Step-by-step explanation:

Solution:

Given:

A, B and C are sets.

We have to prove that:

(A − B) ∩ (C − B) = (A ∩ C) − B.

Use an element x,

Let x Є (A – B), by the definition of difference:

X Є A  and x ∉ B.

And

X Є A ∩ C and x ∉ B

X Є (A ∩ C) – B

Now let x Є C – B

By the definition of difference:

X Є C and x ∉ B

By definition of intersection: x Є C and x Є( A ∩ C)  

By definition of difference, using that x Є(A ∩ C) And x  ∉ B:

X Є ( A ∩ C) – B

We know that: x Є (A – B) ∩ ( C – B)

Is equivalent to x Є A – B  and x Є C – B

So, ( x Є A and x ∉ B)

And (x Є C and x  ∉  B)

From this we know that x Є (A ∩ C) – B.

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step-by-step explanation: