Closed sets - Simple question, if you know the answer.

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First I am in [itex]\mathbb{R}[/itex] with the standard metric [d(x,y)=|y-x|]. Is [itex][0,\infty)[/itex] considered a closed set? I would think yes, since I would consider [itex](-\infty,0)[/itex] to be an open set. However, I can't seem to find any examples like this in our book, and I have yet to be able to find anything online either to clarify this. I guess I am not sure how to deal with infinity. Thoughts? Thanks!


edit... Last time I posted part of a question people wanted to see the whole thing.



So here is the question: Let [itex](\mathbb{R},d)[/itex] be the real line with the standard metric. Give an example of a continuous function [itex]f:\mathbb{R}\to\mathbb{R}[/itex], and a closed set [itex]F\subseteq \mathbb{R}[/itex], such that [itex]f(F) = \{f(x) : x \in F\}[/itex] is not closed.

So I was thinking of taking [itex]f(x)=e^x[/itex] and taking [itex](-\infty,0][/itex] as my closed set. Since that would be mapped into [itex](0,1][/itex] which is not closed. Here arises my question of is [itex](-\infty,0][/itex] closed.

Also, if you have any insightful examples for this question, I would love to see them. Thanks!
 
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  • #2
Dick
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If you consider the complement of the set to be open, then you had better consider the set to be closed. Right?
 
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Well yeah, that is why I was guessing it to be closed. However, I was not sure if either set was open or closed. From your response, though, I am guessing I was right in my guesses.
 
  • #4
Dick
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Please say, "It's not a guess". :smile:
 
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I guess we (I guess I should maybe say I?) used the word guess too much, and I guess I am confused by your response :rofl:

OK, so [itex](0,\infty)[/itex] is open, and hence the complement is closed. And thus my example of a function and closed set produces an open set. :smile:
 
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Dick
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I guess we (I guess I should maybe say I?) used the word guess too much, and I guess I am confused by your response :rofl:

OK, so [itex](0,\infty)[/itex] is open, and hence the complement is closed. And thus my example of a function and closed set produces an open set. :smile:
That's better. Show some CONFIDENCE!
 
  • #7
dextercioby
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First I am in [itex]\mathbb{R}[/itex] with the standard metric [d(x,y)=|y-x|]. Is [itex][0,\infty)[/itex] considered a closed set? I would think yes, since I would consider [itex](-\infty,0)[/itex] to be an open set. I guess I am not sure how to deal with infinity. Thoughts?
Here's something for you. What is the closure/adherence of [itex] [0,\infty) [/itex] ?
 
  • #8
HallsofIvy
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You haven't told us what DEFINITION of closed set you are using! Don't you think that's important?

The reason I mention that is that there are, in fact, several different definitions of "closed" set. One that is commonly used is "complement of an open set". Since you say " I would think yes, since I would consider [itex](-\infty,0)[/itex] to be an open set", I take it you are using that one.

Okay, how would you prove that the complement of [itex][0,\infty)[/itex], which is [itex](-\infty,0)[/itex] to be an open set? Welll, that depends on your definition of open set! Probably you are using "every member is an interior point". So what is the definition of "interior point"? That some neighborhood of the point is a subset of your original set. If x is in [itex](-\infty, 0)[/itex], then is it a negative number. What if you take [itex]\delta[/itex]= -a/2? Can you show that [itex](x- \delta, x+ \delta)[/itex] contains only negative numbers? If so then you have shown that [itex](-\infty, 0)[/itex] is an open set and so [itex][0,\infty)[/itex] is a closed set.

You probably learned, in Calculus or before, that an interval is "open" if it has "(" and ")" and closed if it has "[" and "]". Of course, [itex][0, \infty)[/itex] is confusing because it has both! But [itex]\infty[/itex] isn't a real number so ...

That can be used as another definition of "open" and "closed":
We say that a point is an "interior" point of a set if some neighborhood is a subset of the set (same definition as above). We say that a point is an "exterior" point of a set if it is an interior point of the complement of the set (some neighborhood contains NO member of the set). We say that a point is a "boundary" point of a set if it is neither an interior point or an exterior point of the set.

A set is open if it contains NONE of its boundary points, closed if it contains ALL of its boundary points.
What are the boundary points of [itex][0, \infty)[/itex]?

For extra credit: it is clear that if a set contains some of its boundary points, but not all, it is neither open nor closed. Under what conditions is a set both open and closed?
 
  • #9
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Here's something for you. What is the closure/adherence of [itex] [0,\infty) [/itex] ?
I would say the closure would be still [itex][0, \infty)[/itex]



You haven't told us what DEFINITION of closed set you are using! Don't you think that's important?

The reason I mention that is that there are, in fact, several different definitions of "closed" set. One that is commonly used is "complement of an open set". Since you say " I would think yes, since I would consider [itex](-\infty,0)[/itex] to be an open set", I take it you are using that one.

Okay, how would you prove that the complement of [itex][0,\infty)[/itex], which is [itex](-\infty,0)[/itex] to be an open set? Welll, that depends on your definition of open set! Probably you are using "every member is an interior point". So what is the definition of "interior point"? That some neighborhood of the point is a subset of your original set. If x is in [itex](-\infty, 0)[/itex], then is it a negative number. What if you take [itex]\delta[/itex]= -a/2? Can you show that [itex](x- \delta, x+ \delta)[/itex] contains only negative numbers? If so then you have shown that [itex](-\infty, 0)[/itex] is an open set and so [itex][0,\infty)[/itex] is a closed set.

You probably learned, in Calculus or before, that an interval is "open" if it has "(" and ")" and closed if it has "[" and "]". Of course, [itex][0, \infty)[/itex] is confusing because it has both! But [itex]\infty[/itex] isn't a real number so ...

That can be used as another definition of "open" and "closed":
We say that a point is an "interior" point of a set if some neighborhood is a subset of the set (same definition as above). We say that a point is an "exterior" point of a set if it is an interior point of the complement of the set (some neighborhood contains NO member of the set). We say that a point is a "boundary" point of a set if it is neither an interior point or an exterior point of the set.

A set is open if it contains NONE of its boundary points, closed if it contains ALL of its boundary points.
What are the boundary points of [itex][0, \infty)[/itex]?

For extra credit: it is clear that if a set contains some of its boundary points, but not all, it is neither open nor closed. Under what conditions is a set both open and closed?
Sorry about the definition. Our definition of open was containing none of its boundary points. However, in our last homework we had to prove that a set is open iff its complement is closed, and that a set E is open iff E = Int(E) [Interior of E] (among a ton of other properties).

The boundary point/s of [itex][0,\infty)[/itex] would be 0 which is contained in the set and hence the set is closed.

A set could be open and closed if there were no boundary points. This could happen if the set was empty, or if we had the complete space as our set.
 

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