Open and Closed Set with Compactness

In summary, a closed set S⊂ℝ and a continuous function f: ℝ-->ℝ such that f(S) is not closed can be represented by the function f(x)=e^x and the set S=R. Similarly, for an open set U⊂ℝ and a continuous function f: ℝ-->ℝ such that f(U) is not open, the function f(x)=x^2 and the set U=R can be used as an example.
  • #1
trap101
342
0
Give an example of:

a) a closed set S[itex]\subset[/itex]ℝ and a continuous function f: ℝ-->ℝ such that f(S) is not closed;

b) an open set U[itex]\subset[/itex]ℝ and a continuous function f: ℝ-->ℝ such that f(U) is not open


Solution:

a) e^x b) x^2

here's my problem, this is what was given in the solutions, but I don't see how the two respective answers are an open and closed set. The continuity isn't a problem, but the open and close concept is bugging me. How is e^x closed if I can pick any value of "x"? Like my interpretation of closed was any stated interval where you may have to say [a,b] is the interval also topologically I viewed it as the closure = union of the interior points an boundary points.
 
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  • #2
What is the closure of ℝ? Thus, ℝ is closed. If f(x)=e^x, what is f(ℝ)?

ℝ is also open. If f(x)=x^2, what is f(ℝ)?



What do you mean when you say e^x is closed?
 
  • #3
algebrat said:
What is the closure of ℝ? Thus, ℝ is closed. If f(x)=e^x, what is f(ℝ)?

ℝ is also open. If f(x)=x^2, what is f(ℝ)?



What do you mean when you say e^x is closed?



I'm going to hate myself because it's probably really simple, but I think it may be that I don't even understand the question.

So the closure of ℝ has to be ℝ itself. as for f(ℝ) i take it you mean I'm just taking ℝ and mapping it its image in this case f(ℝ) = ℝ. But then if I'm right in what your saying, all it is, is that I'm taking a number (any number) in ℝ and applying e^x to it. But how does it show "not open"
 
  • #4
The "answers" that they gave you do not, in fact, answer the questions that were asked, so you are justified in your confusion. The original questions asks for a set and a function, and the answer key only gives you a function.
So that we are clear on notation, the image of a set X under a function f is f(X): the set of all elements y such that y = f(x) for some x in X.
For the first question, you need a set S whose image f(S) is not closed for a specific continuous function f. The answer key claims that the function f(x) = e^x gives us this property for some set S, so it is still up to us to provide that set. If we use S = R, then we see that f(R) is the set of all positive real numbers, not R. The exponential function is not 0 for any real number and is never negative. Is the set of all positive real numbers closed relative to R?
An easy way to check this is to see if the set contains its closure, as a set is closed if and only if it contains its closure.
If f(R) were R, we could not use it, because R is closed, and we are looking for an image that is not closed.
The second function maps U = R to the set of all non-negative real numbers. Is this set open relative to R? Why or why not?
 
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  • #5
slider142 said:
The "answers" that they gave you do not, in fact, answer the questions that were asked, so you are justified in your confusion. The original questions asks for a set and a function, and the answer key only gives you a function.
So that we are clear on notation, the image of a set X under a function f is f(X): the set of all elements y such that y = f(x) for some x in X.
For the first question, you need a set S whose image f(S) is not closed for a specific continuous function f. The answer key claims that the function f(x) = e^x gives us this property for some set S, so it is still up to us to provide that set. If we use S = R, then we see that f(R) is the set of all positive real numbers, not R. The exponential function is not 0 for any real number and is never negative. Is the set of all positive real numbers closed relative to R?
An easy way to check this is to see if the set contains its closure, as a set is closed if and only if it contains its closure.
If f(R) were R, we could not use it, because R is closed, and we are looking for an image that is not closed.
The second function maps U = R to the set of all non-negative real numbers. Is this set open relative to R? Why or why not?
Ok so from your explanation then since the exponential function can never be 0 or negative this means that the image of ℝ i.e f(ℝ) will not contain the "whole" group of Real Numbers since 0 and the negatives are included in this group as such it cannot be closed.

But from what I just said, that would mean that in the second question f(U) would also have to be open because it does not include the negative Real numbers either after the function, x^2, is applied to it.
 
  • #6
trap101 said:
Ok so from your explanation then since the exponential function can never be 0 or negative this means that the image of ℝ i.e f(ℝ) will not contain the "whole" group of Real Numbers since 0 and the negatives are included in this group as such it cannot be closed.
No... The interval [0,1] is closed, but it obviously doesn't contain all of R. What does it mean when you say a set is open or closed? Once you review that, you can answer the following: The function ex maps R to (0, ∞). Is that interval closed?
 
  • #7
vela said:
No... The interval [0,1] is closed, but it obviously doesn't contain all of R. What does it mean when you say a set is open or closed? Once you review that, you can answer the following: The function ex maps R to (0, ∞). Is that interval closed?



Set open = the set contains all of its interior points only

Set Closed = the set contains its interior points and its boundary points.


Oh...since ex can never be 0 that means the boundary point "0" is not a part of the set. Therefore f(S) cannot be closed. Since S = ℝ is closed applying the function to S gives f(S), but since we picked our function to be ex, the value 0 is not present in f(s) which we would need for it to be closed.


in the second case x2 maps to the [0,∞] because every value of the reals can be obtained i.e, the boundary points. So then this means it's closed.
 
  • #8
trap101 said:
Set open = the set contains all of its interior points only

Set Closed = the set contains its interior points and its boundary points.Oh...since ex can never be 0 that means the boundary point "0" is not a part of the set. Therefore f(S) cannot be closed. Since S = ℝ is closed applying the function to S gives f(S), but since we picked our function to be ex, the value 0 is not present in f(s) which we would need for it to be closed.

Your proof would be more complete if you completely describe the image, not just mention some point that is not in the image. But It looks like you have the idea.
in the second case x2 maps to the [0,∞] because every value of the reals can be obtained i.e, the boundary points. So then this means it's closed.

The notation is [0,∞).
 
  • #9
algebrat said:
Your proof would be more complete if you completely describe the image, not just mention some point that is not in the image. But It looks like you have the idea.




The notation is [0,∞).




In the first one what am I missing to describe the image? and thanks everybody for the help.
 
  • #10
trap101 said:
the value 0 is not present in f(s) which we would need for it to be closed.
In the first one what am I missing to describe the image? and thanks everybody for the help.

Simply saying 0 is not in f(S) does not suffice. The empty set does not contain 0 but it is closed. You need to explicitly say what f(S) is.

And you're welcome, and I would like to thank everyone also, for being there to expand when I was not sure where to start.
 

1. What is an open set?

An open set is a set in a topological space where every point in the set has a neighborhood contained entirely within the set. In other words, for any point in an open set, there exists a small enough open ball around that point that is also contained within the set.

2. What is a closed set?

A closed set is a set in a topological space that contains all of its limit points. In other words, if a sequence of points within the set converges to a point outside the set, then the set is not closed. A closed set can also be thought of as the complement of an open set.

3. What is compactness?

Compactness is a property of a topological space where every open cover of the space has a finite subcover. In simpler terms, this means that for any collection of open sets that cover the space, there exists a finite number of open sets within that collection that also cover the space. This is often referred to as the "finite subcover property."

4. How does compactness relate to open and closed sets?

In general, compactness is a stronger property than either being open or closed. However, there are some connections between compactness and open/closed sets. For example, in a Hausdorff space, a set is compact if and only if it is both closed and bounded. Additionally, a subset of a compact set is compact if and only if it is closed.

5. Why is compactness important in mathematics?

Compactness is an important concept in mathematics because it allows us to make statements about infinite sets or spaces using only finitely many elements or pieces. This makes it a powerful tool for proving theorems and solving problems in various branches of mathematics, such as analysis, topology, and geometry. Compactness also has applications in fields such as physics and computer science.

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