anonymous
  • anonymous
help needed here please
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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.
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • Michele_Laino
I can solve that question using the Theorem of Category of Baire

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anonymous
  • anonymous
yes please use that theorem
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • 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
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
now, let: \[\Large {O_n} = {\left\{ {{V_n}} \right\}_{n \in \mathbb{N}}}\] be our countable collection
anonymous
  • anonymous
ok please continue , i am here
Michele_Laino
  • Michele_Laino
the subsequent set: \[\Large \overline {{ \cap _n}{V_n}} = X\]
Michele_Laino
  • Michele_Laino
since by the Baire's theorem the set: \[\Large {{ \cap _n}{V_n}}\] is dense in X
anonymous
  • anonymous
is that all?
Michele_Laino
  • Michele_Laino
no, please let's suppose, by absurdum, that On is empty
Michele_Laino
  • Michele_Laino
that means each Vn is empty: \[\Large {O_n} = \emptyset \Rightarrow {V_n} = \emptyset ,\quad \forall n\]
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
so, also the intersection is empty: \[\Large { \cap _n}{V_n} = \emptyset \]
Michele_Laino
  • Michele_Laino
because the intersection of empty sets is also empty
anonymous
  • anonymous
yap
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
that's all!
anonymous
  • anonymous
thanks . want to copy it and study but can i ask ine last question for now?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • 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.
Michele_Laino
  • Michele_Laino
please I have to go out now, I will come back between 2 hours
anonymous
  • anonymous
@Michele_Laino
Michele_Laino
  • 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: \[\Large \mathop {\lim }\limits_{n \to \infty } f\left( {{x_n}} \right) = f\left( {{x_0}} \right)\]
Michele_Laino
  • 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
anonymous
  • anonymous
ok
anonymous
  • anonymous
is that all?
Michele_Laino
  • Michele_Laino
yes! we are done
anonymous
  • anonymous
waw, thanks .
Michele_Laino
  • Michele_Laino
:)

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