## anonymous one year ago help needed here please

1. anonymous

Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty.

2. anonymous

@Michele_Laino

3. Michele_Laino

I can solve that question using the Theorem of Category of Baire

4. anonymous

yes please use that theorem

5. Michele_Laino

ok!

6. Michele_Laino

Using that theorem, we can state this: since X is a complete metric space, then any countable collection of dense open subsets has dense intersection

7. anonymous

ok

8. Michele_Laino

now, let: $\Large {O_n} = {\left\{ {{V_n}} \right\}_{n \in \mathbb{N}}}$ be our countable collection

9. anonymous

ok please continue , i am here

10. Michele_Laino

the subsequent set: $\Large \overline {{ \cap _n}{V_n}} = X$

11. Michele_Laino

since by the Baire's theorem the set: $\Large {{ \cap _n}{V_n}}$ is dense in X

12. anonymous

is that all?

13. Michele_Laino

no, please let's suppose, by absurdum, that On is empty

14. Michele_Laino

that means each Vn is empty: $\Large {O_n} = \emptyset \Rightarrow {V_n} = \emptyset ,\quad \forall n$

15. anonymous

ok

16. Michele_Laino

so, also the intersection is empty: $\Large { \cap _n}{V_n} = \emptyset$

17. Michele_Laino

because the intersection of empty sets is also empty

18. anonymous

yap

19. Michele_Laino

now, the empty set is both a closed and open set, and we can write this: $\Large \emptyset = \overline \emptyset = \overline {{ \cap _n}{V_n}} = X$ which is a contradiction, since by hypothesis, X is not empty

20. Michele_Laino

that's all!

21. anonymous

thanks . want to copy it and study but can i ask ine last question for now?

22. Michele_Laino

ok!

23. anonymous

Let (X, d) and (Y, d) be metric spaces and f a mapping of X into Y. Let τ1 and τ2 be the topologies determined by d and d1 respectively. Then f(X, τ) (y, τ) is continuous if and only if ; that is if x1, x2, . . . , xn, . . . , is a sequence of points in (X, d) converging to x, show that the sequence of points f(x1), f(x2), . . . , f(xn), . . . in (Y, d) converges to x.

24. Michele_Laino

please I have to go out now, I will come back between 2 hours

25. anonymous

@Michele_Laino

26. Michele_Laino

a proof can be this: let's suppose that f is contnuous, then the sequence x1, x2,...xn can be viewed as a continuous function from the set of natural number to the metric space (X,d), therefore, using the theorem of composition of continuous function, we can say that the function: $\Large \mathbb{N} \to Y:n \to f\left( {{x_n}} \right)$ is also continuous and we have: $\Large \mathop {\lim }\limits_{n \to \infty } f\left( {{x_n}} \right) = f\left( {{x_0}} \right)$

27. Michele_Laino

vice versa, let's suppose that f is not a continuous function at x_0. Then we can find a neighborhood V around f(x_0) of Y, such that for each neighborhood U around x_0 of X we have: $\Large f\left( U \right) \not\subset V$ We can choose each of Us like a ball of radius r= 1/n, n=1,2,3,.... Inside each of such balls U, we can find an element x_n such that: $\Large f\left( {{x_n}} \right) \notin V$ in so doing we got a subsequence x1, x2,...,xn which converges to x_0, nevertheless f(x_n) doesn't converge to f(x_0), namely a contradiction

28. anonymous

ok

29. anonymous

is that all?

30. Michele_Laino

yes! we are done

31. anonymous

waw, thanks .

32. Michele_Laino

:)