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I have been sick for my real analysis class and I am a having a tough time getting the book.

Could someone explain the what open, closed and compact sets are and how they are related with uniform continuity? I think the hinging point was a poor understanding of accumulation points.

Thanks

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Your question is very broad and a little vague. What is it you're having trouble with? Have you read these? en.wikipedia.org/wiki/Compact_space, en.wikipedia.org/wiki/Uniformly_continuous, en.wikipedia.org/wiki/Open_set, en.wikipedia.org/wiki/Closed_set –  Jonas Meyer Nov 12 '10 at 6:09
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sorry! i almost read the 'for' in the first sentence as 'of' –  user1709 Nov 12 '10 at 9:46
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3 Answers 3

up vote 8 down vote accepted

I assume you're talking about open, closed, and compact subsets of a metric space. If you're just talking about the real line $\mathbb{R}$, then ignore the commentary about abstract metric spaces and just replace the term "neighborhood" with "open interval."


In metric spaces, a set $O$ is open if every point in $O$ is an interior point of $O$. In other words, every point of $O$ can be surrounded by an open ball (neighborhood) which itself lies in $O$. So, for instance, the open intervals $(a,b)$ are open subsets of $\mathbb{R}$, and discs $\{x^2 + y^2 \lt R^2 \}$ are open subsets of $\mathbb{R}^2$.

However, one thing to keep in mind is that in more abstract metric spaces, the open sets can be a little strange-looking. For instance, there are metric spaces where every subset is an open set!


Now, a set $F$ is closed if every accumulation point of $F$ is in $F$. Accumulation points are points $x$ such that every open neighborhood around $x$ must intersect the set in question. So, for instance, the accumulation points of the open disc $\{x^2 + y^2 \lt R^2\}$ are precisely the set of points in the closed disc $\{x^2 + y^2 \leq R^2\}$.

One way to think about closed sets is that they are closed under the operation of taking limits. That is, if one were to take a sequence of points in a closed set, then their limit would still have to be in the set. (This can be made rigorous and should be proven at some point.)

As with open sets, the closed sets of a metric space can still be rather strange-looking, and again, there are examples of spaces in which every subset is closed. Nevertheless, it is always true that finite subsets of metric spaces are closed.

You should note also that if $(X,d)$ is a metric space, then the sets $\emptyset$ and $X$ are always going to be both open and closed. In general, the subsets of a metric space may be open, closed, both or neither. (So just because a set is not open does not make it closed!)


Compact sets are a little less intuitive. The rigorous definition is that a set $K$ is compact if every open cover of $K$ has a finite subcover. Since I don't want to babble for too long, I won't go into what this means in detail. However, I've always intuited compact sets as being "super closed" sets -- they're like closed sets, but even stronger.

It is a fact (whose proof you should learn or create) that every compact subset of a metric space is closed and bounded, but the converse is not necessarily true. In the euclidean spaces $\mathbb{R}^n$, the converse is, in fact, true, but this is not obvious, and is known as the Heine-Borel Theorem.

Unlike open and closed sets, in a metric space $(X,d)$, the set $X$ may not be compact. Therefore, it makes sense to talk about "compact metric spaces," whereas it doesn't make sense to talk about "open" or "closed" metric spaces (since they're all open and closed subsets of themselves).


Uniform continuity is a stronger condition than continuity. That is, every uniformly continuous function is continuous, but not necessarily conversely. It is a theorem (Heine-Cantor) that on compact sets, the converse will hold. So on compact sets, continuity is equivalent to uniform continuity. Again, this is an important theorem whose proof you should learn.

Essentially, uniform continuity of a function $f$ on a set $E$ can be thought of as saying that the "degree of continuity" is the same for every point $x$ in the set $E$. This can be made slightly more precise by saying that the $\delta > 0$ that one gets in the $\epsilon-\delta$ definition of continuity can be chosen to be the same for all $x$ in $E$ -- that is, it will depend only on $\epsilon$, and not on the point $x$. For plain old continuous functions, the $\delta$ will depend on both $\epsilon$ and $x$.

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Not sure why the curly braces aren't displaying... if someone knows how to fix that, please do edit accordingly. –  Jesse Madnick Nov 12 '10 at 6:30
    
It was the less than symbols, I think. I changed them to \lt. –  Jonas Meyer Nov 12 '10 at 6:34
    
Ah, thank you, Jonas. Much better now. –  Jesse Madnick Nov 12 '10 at 6:40
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So I'm going to try to explain what you asked for, though you should probably get your book ASAP or reference a free analysis book and Wikipedia, and probably most importantly, get someone to take notes for you in your class.

A point p is an accumulation/limit point of a set E if every neighborhood around that point has an element of the set E not equal to p. Think of a neighborhood as an ball around a point. A neighborhood in $\mathbb{R}$ (the real line) is just an interval.

A set is closed if every limit point of a set E is inside E. An example of where you could have a limit point that is not an element of a set is given by the set $\mathbb{Q}$ (the rationals). You can build a series of rational numbers that converge to say for example $\pi$, and hence $\pi$ is a limit point of $\mathbb{Q}$.

That is to say if you have an any interval around $\pi$, you'll have an element of $\mathbb{Q}$. However, pi is not an element of $\mathbb{Q}$, hence the rationals aren't a closed set (Actually every irrational number is a limit point of $\mathbb{Q}$... you may want to think about why).

A set is open if every point of that set is an interior point. A point is an interior point if there exists at least one open ball around that point such that that open ball is contained entirely in the set.

For your usage, assuming that you'll be studying $\mathbb{R}$, a compact set is one which is closed and bounded. A set is bounded if you can but an open ball at its center that covers the entire set. Think [-1,1], with the ball centered at 0, and radius 1. Note that (-1,1) is not closed, since both endpoints are limit points of the set, but not elements of the set (ie. you can have an infinite series that converges to 1, and has elements that are in (-1,1), meaning 1 is a limit point, yet 1 is not an element of the set).

The relationship between compactness and uniform continuity is given by Heine-Cantor. Hope that helps.

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It looks like we all wrote these answers at about the same time. Hopefully this serves as an alternative reference. –  Rishi Nov 12 '10 at 21:06
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Difficult one to answer, given the numerous definitions of open sets. Also - more information on what you know, etc, would be useful.

I'll try not to be overly formal - someone will probably be more so than me.

Take a point $x$ in a set. If for every point $x$ we can find another point arbitrarily close then this is an open set. From the standard wikipedia definition:

A subset $U$ of the Euclidean n-space $\mathbb{R}^n$ is called open if, given any point $x$ in $U$, there exists a real number $\epsilon > 0$ such that, given any point $y$ in $\mathbb{R}^n$ whose Euclidean distance from $x$ is smaller than $\epsilon$, $y$ also belongs to $U$.

Now (again see Wikipedia) you can replace the Euclidean space with any metric space, and come up with an equivalent definition.

Closed sets are sets where the complement is open. One of the most confusing aspects is there exists sets that are both open and closed. These are known as Clopen sets. The definition of closed sets you are after may be the one involving accumulation points.

Definition: $x$ is an accumulation points of $U$ if every $\epsilon$ neighborhood of $x$ contains a points ($\ne x$) in $U$.

A subset of $U$ is closed if if it contains all its accumulation points.

I see Jesse has posted a nice answer - I will leave this as an alternative reference.

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