Simple Question on continuity

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Homework Statement



1) If f is a continuous mapping from a matric space X to metric space Y. A E is a subset of X.
The prove that f(closure(E)) subset of closure of f(E).

2) Give an example where f(closure (E)) is a proper subset of closure of f(E).


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The Attempt at a Solution




My problem is the second part, although I am unsure of my solution to part (1) as well.

If E is closed then E= closure of E. Hence, for all x that belong to E , f(x) belong to Y. Hence, f(closure of E) = f(E) = set of all f(x). Now, the closure of the set of all f(x) shall be f(x) (as collection of f(x) in Y is also closed).

Hence, the closure of f(E)= f(closure of (E)).

Now, if E is open, then closure of E = E U E', where E' contains limit points. Now, f(closure of E) = f(E) and the set of limit points that are not in E.
Where as f(E) shall be f(x) .

(Kind of lost here).

Also, can't think of a possible example that satisfies (2). I was thinking f(x) = 1/x , where x = (0,infinity). But this does not follow proper subset example.

Please help.
 

Answers and Replies

  • #2
HallsofIvy
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Exactly what is your definition of "continuous". Most commonly used is "f: X->Y is continuous if and only if f-1(B) is open, in X, for every open set B in Y." In that case, you might want to look at the complements of "closure of E" and "closure of f(E)" which are then open sets. Do you see that f-1(complement of closure (E)) is an open set?

Sometimes we define f(x) to be continuous more in keeping with the Calculus I definition: "f is continuous on X if and only if for each a in X, if {an} is a sequence of points in X, converging to a, then {f(xn} is a sequence of points in Y converging to f(a)."

Of course, for every a in the closure of E, there must exist a sequence of points in E converging to a.
 

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