Question

help needed here please

Answer 1

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

Answer 2

@Michele_Laino

Answer 3

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

Answer 4

yes please use that theorem

Answer 5

ok!

Answer 6

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

Answer 7

ok

Answer 8

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

Answer 9

ok please continue , i am here

Answer 10

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

Answer 11

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

Answer 12

is that all?

Answer 13

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

Answer 14

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

Answer 15

ok

Answer 16

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

Answer 17

because the intersection of empty sets is also empty

Answer 18

yap

Answer 19

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

Answer 20

that's all!

Answer 21

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

Answer 22

ok!

Answer 23

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.

Answer 24

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

Answer 25

@Michele_Laino

Answer 26

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:
\[\Large \mathop {\lim }\limits_{n \to \infty } f\left( {{x_n}} \right) = f\left( {{x_0}} \right)\]

Answer 27

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

Answer 28

ok

Answer 29

is that all?

Answer 30

yes! we are done

Answer 31

waw, thanks .

Answer 32

:)