Question about continuous function

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Homework Statement


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?


Homework Equations





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:)
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


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?


Homework Equations





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:)
?? 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!
 
  • #3
Dick
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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!

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.
 
  • #4
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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 :)
 
  • #5
Dick
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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 :)

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

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