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anonymous
 one year ago
help needed here please
anonymous
 one year ago
help needed here please

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that On is not empty.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I can solve that question using the Theorem of Category of Baire

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes please use that theorem

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Using that theorem, we can state this: since X is a complete metric space, then any countable collection of dense open subsets has dense intersection

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, let: \[\Large {O_n} = {\left\{ {{V_n}} \right\}_{n \in \mathbb{N}}}\] be our countable collection

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok please continue , i am here

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the subsequent set: \[\Large \overline {{ \cap _n}{V_n}} = X\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since by the Baire's theorem the set: \[\Large {{ \cap _n}{V_n}}\] is dense in X

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, please let's suppose, by absurdum, that On is empty

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that means each Vn is empty: \[\Large {O_n} = \emptyset \Rightarrow {V_n} = \emptyset ,\quad \forall n\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, also the intersection is empty: \[\Large { \cap _n}{V_n} = \emptyset \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1because the intersection of empty sets is also empty

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks . want to copy it and study but can i ask ine last question for now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let (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.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please I have to go out now, I will come back between 2 hours

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1a 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: \[\Large \mathop {\lim }\limits_{n \to \infty } f\left( {{x_n}} \right) = f\left( {{x_0}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1vice 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we are done
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