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I have this exercise and I don't know how to solve it Let $(A_n)_n \subset (E,d),(B_n)\subset (F,d')$ 2 sequence of sets from a metric space,and $f: E \rightarrow F$ a continuous aplication .

Prove that :

1)$\displaystyle f(\limsup_{n\to\infty} A_n) \subset \limsup_{n\to\infty} f(A_n)$

2)$\displaystyle f(\liminf_{n\to\infty} A_n) ‎\subset‎ \liminf_{n\to\infty} f(A_n)$

3)$\displaystyle \limsup_{n\to\infty}f^{-1}(B_n) \subset f^{-1}(\limsup_{n\to\infty} B_n)$

4)$\displaystyle \liminf_{n\to\infty}f^{-1}(B_n) \subset f^{-1}(\liminf_{n\to\infty} B_n)$

with: $x\in \overline\lim(A_n) \Rightarrow$ $ x\in \displaystyle\bigcap_{\varepsilon>0}\bigcap_{N>0}\bigcup_{n\geq N} (A_n)_\varepsilon \Rightarrow \forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(x,A_n)< \varepsilon$

help me please , thank you .

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    $\begingroup$ Metrics and continuity are irrelevant here, this is a purely set-theoretical result. // What did you try? $\endgroup$ – Did Feb 13 '13 at 18:01
  • $\begingroup$ @Did I believe the continuity is relevant. See en.wikipedia.org/wiki/… for definitions. $\endgroup$ – ferson2020 Feb 13 '13 at 18:33
  • $\begingroup$ @ferson2020 Now that the OP modified their post, continuity is involved. Before that, a quite usual interpretation was that it was not. $\endgroup$ – Did Feb 13 '13 at 19:42
  • $\begingroup$ 1) i can say: $x\in f(\limsup A_n) \Rightarrow \exists y\in \limsup A_n , x=f(y) \Rightarrow \forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(y,A_n)< \varepsilon ,x=f(y)$ as f is continuous then $\forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(f(y),f(A_n))< \varepsilon$ then $x=f(y) \in \limsup f(A_n)$ , it's correct ? please $\endgroup$ – Vrouvrou Feb 14 '13 at 11:55
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Hint: See, Own Lecture Notes Functional Analysis. Theorem 2.15.1, p17.

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  • $\begingroup$ See , this is for multivalued set :$x\in f(\limsup A_n) \Rightarrow \exists y\in \limsup A_n , x=f(y)$ ,y\in \limsup A_n \Rightarrow \liminf d(y,A_n)=0$ $\endgroup$ – Vrouvrou Feb 13 '13 at 19:16
  • $\begingroup$ 1) i can say :$x\in f(\limsup A_n) \Rightarrow \exists y\in \limsup A_n , x=f(y) \Rightarrow \forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(y,A_n)< \varepsilon ,x=f(y) $ as $f$ is continuous then $\forall \varepsilon >0, \forall N>0, \exists n\geq N ;d(f(y),f(A_n))< \varepsilon $ then $x=f(y) \in \limsup f(A_n)$ , it's correct ? $\endgroup$ – Vrouvrou Feb 14 '13 at 9:52
  • $\begingroup$ I edited my post because the last refrence is about sequence of numbers but this is about sequence of sets. see Theorem 2.15.1, p17 with usual definition of limSup and liminf, continuous of $f$ is irrelevant, but with your definition i am not sure, that what you write is true. $\endgroup$ – M.Sina Feb 14 '13 at 12:45

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