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Question about continuous function

  1. Aug 6, 2008 #1
    1. The problem statement, all variables and given/known data
    If f is a continuous mapping of a metric space X into a metric space Y, Let E be any subset of X. How to show, by an example, that f([tex]\overline{E}[/tex]) ([tex]\overline{E}[/tex] is the closure of E) can be a proper subset of [tex]\overline{f(E)}[/tex] ? And is there something wrong with my attempt below?

    2. Relevant equations

    3. The attempt at a solution
    If E is compact, [tex]\overline{E}[/tex] = E, f(E) is compact, [tex]\overline{f(E)}[/tex] = f(E). Hence, f([tex]\overline{E}[/tex])= [tex]\overline{f(E)}[/tex]
    If E is not compact, [tex]\overline{E}[/tex] is closed and hence is compact, if E is bounded in R^{k}. f([tex]\overline{E}[/tex]) is compact and hence [tex]\overline{f(\overline{E})}[/tex] = f([tex]\overline{E}[/tex]).
    since f(E) [tex]\subset{f(\overline{E})}[/tex] , [tex]\overline{f(E)}[/tex] [tex]\subset{\overline{f(\oveline{\overline{E}})}} [/tex] = f([tex]\overline{E}[/tex]). It is also true that f([tex]\overline{E}[/tex]) [tex]\subset{\overline{f(E)}}[/tex]. Hence f([tex]\overline{E}[/tex]) = [tex]\overline{f(E)}[/tex]
    In both cases, f([tex]\overline{E}[/tex]) is not a proper subset of [tex]\overline{f(E)}[/tex]

    I've no idea of other kind of function that is continuous and with E otherwise defined.
    Any hint would be greatly appreciated:)
  2. jcsd
  3. Aug 6, 2008 #2


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    ?? The problem was to show that f([tex]\overline{E}[/tex]) can be a proper subset of [tex]\overline{f(E)}[/tex] and you have given two examples where that is NOT true. What's "wrong with that" is that you haven't answered the question!
  4. Aug 6, 2008 #3


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    Exactly. You want an example. There are perfectly fine examples from R->R. You don't need anything exotic. Can you find one? Hint, think about what happens if f has a horizontal asymptote.
  5. Aug 6, 2008 #4
    to HallsofIvy,
    I gave those two examples because I could not find a disired example :(

    to Dick,
    O I think I get it now...Let E=[1,[tex]\infty[/tex]), f(x)=1/x and everything is done.....it is strange that I didn't get it...

    Thanks a lot to all :)
  6. Aug 6, 2008 #5


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    Not too strange. You were only thinking of bounded sets. This is called "thinking in the box". You're welcome.
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