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Topological continuity (a few questions).

  1. Jan 16, 2008 #1


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    1.suppose that f:X->Y is continuous. if x is a limit point of the subset A of X, is it necessarily true that f(x) is a limit point of f(A)?
    2. suppose that f:R->R is continuous from the right, show that f is continuous when considered as a function from R_l to R, where R_l is R in the lower limit topology. (munkres' notation).

    Now for 1, I think the answer is no, but I don't find a counterexample if someone could give me a hint on this ( I'm pretty sure it's easy (-: ), but what I did find is that if f is injective then the answer is yes (I proved it by ad absurdum), so my hunch a counterexample should be with a function which is not injective.

    Now for two it seems easy enough, if V is open in R, then it contains an open interval, let it be (a,b), now then f^-1(V) contains f^-1((a,b))={x in R_l|f(x) in (a,b)}
    now i need to prove that f^-1((a,b)) is an interval of the form: [x0,x1), but im struggling with that.
    any hints?

    thanks in advance.
  2. jcsd
  3. Jan 16, 2008 #2
    Are you given any special topology on the sets X,Y?

    You could start by assuming that f(x) is not a limit point of f(A), i.e. there exists a neighborhood U of f(x) in Y, such that...
  4. Jan 16, 2008 #3


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    1. How about taking a constant function from R->R? This looks like it works, provided I'm not mixing up the terminology.

    2. f being continuous doesn't mean it pulls back basic sets to basic sets, only to open sets.
  5. Jan 16, 2008 #4


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    the fact that the answer to 1. is yes, is actually the definition of continuity.

    oops, it depends on your definition of "limit point of", but it is true if it means "in the closure of".

    i realize however that older books define limit points in a strange way.
    Last edited: Jan 16, 2008
  6. Jan 17, 2008 #5


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    well the definition of x a limit point of a subset A of X, is that for every neighbourhood U of x, the intersection between U and A contanis a point which is different than x.

    Now what I did for 1, is that i assumed that f(x) isn't a limit point of f(A) i.e that there exists U open in Y s.t the intersection between f(A) and U is f(x), now i use the fact that
    f^-1(f(A)nU)=Anf^-1(U)=(1), by continuity, f^-1(U) is open in X, and if (1) were equal {x} alone that will be a contradiction for x being a limit point, but (1) also equals: f^-1({f(x)})={x' in X|f(x')=f(x)}, and that would be equal {x} if f is injective as i said, or am i missing something here?

    any hints on question number 2, morphism's hint doesnt help much.
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