Hello forumites I got another question (I have a lots of them, maybe

[_DJ_british_?]

Hello forumites! I got another question (I have a lots of them, maybe I should change my textbook haha).

Suppose you have a continuous real-valued function f on the metric space X and a closed set E in X. Is f(E) closed in the reals? By definition, I know that if f(A) is closed, so is the inverse image of f(A) for a set A in X if f is continuous. But if we know that E is closed, would it be correct to say that f(E) is too?

Thanks a lot!

 

[some_dude]

No, let X = [0, 1), E = X, and f:X->R be the identity function. Then E is closed, but f(E) = [0, 1) as a subset of R is not closed.

 

[_DJ_british_?]

No, let X = [0, 1), E = X, and f:X->R be the identity function. Then E is closed, but f(E) = [0, 1) as a subset of R is not closed.

I could be wrong, but suppose the sequence 0.9, 0.99, 0.999, etc. Then this is a sequence of elements of E that does not converge to an element of E. Thus E isn't closed.

 

[_DJ_british_?]

Let me try something (and I'm surely wrong again) : Suppose a continuous real-valued function on a metric space X and a closed set E in X. Then there is a set V in the reals such that V=f(E) (Because f maps every points in X). If V is open, so is the inverse image of V, that is, E. But E is closed so V must be closed.

 

[Tedjn]

You are forgetting that sets can be neither open nor closed. Take this as your counterexample. X = R. E = [1,∞). f(x) = 1/x. Then, f(E) = (0,1] is not closed. However, it is true that the image of a compact set over a continuous function is compact.

 

[some_dude]

I could be wrong, but suppose the sequence 0.9, 0.99, 0.999, etc. Then this is a sequence of elements of E that does not converge to an element of E. Thus E isn't closed.

No, that is "completeness" you are describing.

"Closed" just requires for any *convergent* sequence in X such that every element of the sequence is in E, it's limit must also be in E. But in the example I gave 1 was not in the metric space [0, 1), therefore that's not a convergent sequence. (Similar to your example, the sequence 9, 99, 999, 9999, ... does not converge in the reals, but the entire real line is still a closed set).

 

[_DJ_british_?]

No, that is "completeness" you are describing.

"Closed" just requires for any *convergent* sequence in X such that every element of the sequence is in E, it's limit must also be in E. But in the example I gave 1 was not in the metric space [0, 1), therefore that's not a convergent sequence. (Similar to your example, the sequence 9, 99, 999, 9999, ... does not converge in the reals, but the entire real line is still a closed set).

Isn't that's weird? That the metric space is't closed but that a subset with the exact same points is? Is it because we define "closure" in respect to a specific metric space instead of, for exemple, the reals (in our case)?

 

[_DJ_british_?]

You are forgetting that sets can be neither open nor closed. Take this as your counterexample. X = R. E = [1,∞). f(x) = 1/x. Then, f(E) = (0,1] is not closed. However, it is true that the image of a compact set over a continuous function is compact.

Well in your exemple a "neither open nor closed" is mapped into a "neither open nor closed" set. But I'm talking about a mapping from a closed set into another one (where we don't know if it is open, closed or neither). I'm sorry if I just didn't understand where you were trying to get!

 

[some_dude]

Isn't that's weird? That the metric space is't closed but that a subset with the exact same points is? Is it because we define "closure" in respect to a specific metric space instead of, for exemple, the reals (in our case)?

No! The metric space [0, 1) is closed! It's the entire space, so it must be open AND closed. You can also see this by applying any of the definitions of closure in a metric space.

 

[Tedjn]

Regarding some_dude's example: I didn't even notice that. Most of the analysis I've done deals with complete metric spaces, so I took it for granted. But the example is valid. As for my example, the set [1,∞) is closed in R.

 

[some_dude]

_DJ_british_?, try playing around with metrics in more generality, you're taking for granted the stuff that holds for the reals holds everywhere.

I could have also given you the counter example:

(R, d) metric space where d(x, y) = 1 if x = y, and d(x, y) = 0 otherwise. Then let f be the identity function. Then f( (-1, 1) ) = (-1, 1), an open and NOT closed set, but (-1, 1) is closed (and open) with respect to the metric I gave above.

 

[_DJ_british_?]

Gah! The last three posts were very insightful! That cleared that up for me. Many thanks guy! And yeah, I may have a bit of difficulty trying to see the reals only as a specific metric space, since I've used them for so long. I'll get used to it, I guess. Anyway, thank you guys.

 

[Landau]

Openness and closedness are actually relative concepts.
[0,1) is closed in [0,1), but not closed in R, nor in [0,1].
Every metric (topological) space is both closed and open in itself.