Topology
It is useful to define a so-called topology. That is we define closed and open sets in a metric space. Before doing so, let us define two special sets.
Let be a metric space, and . Then define the open ball or simply ball of radius around as Similarly we define the closed ball as
When we are dealing with different metric spaces, it is sometimes convenient to emphasize which metric space the ball is in. We do this by writing or .
Take the metric space with the standard metric. For , and we get
Be careful when working on a subspace. Suppose we take the metric space as a subspace of . Then in we get This is of course different from \(B_
ParseError: invalid DekiScript (click for details)Callstack:
at (Bookshelves/Analysis/Introduction_to_Real_Analysis_(Lebl)/08:_Metric_Spaces/8.02:_Open_and_Closed_Sets), /content/body/div[1]/p[5]/span, line 1, column 1
(0,\nicefrac{1}{2}) = (-\nicefrac{1}{2},\nicefrac{1}{2})\). The important thing to keep in mind is which metric space we are working in.
Let be a metric space. A set is open if for every , there exists a such that . See . A set is closed if the complement is open. When the ambient space is not clear from context we say is open in and is closed in .
If and is open, then we say that is an open neighborhood of (or sometimes just neighborhood).
Intuitively, an open set is a set that does not include its “boundary.” Note that not every set is either open or closed, in fact generally most subsets are neither.
The set is neither open nor closed. First, every ball in around , contains negative numbers and hence is not contained in and so is not open. Second, every ball in around , contains numbers strictly less than 1 and greater than 0 (e.g. as long as ). Thus is not open, and so is not closed.
[prop:topology:open] Let be a metric space.
- [topology:openi] and are open in .
- [topology:openii] If are open then is also open. That is, finite intersection of open sets is open.
- [topology:openiii] If is an arbitrary collection of open sets, then is also open. That is, union of open sets is open.
Note that the index set in [topology:openiii] is arbitrarily large. By we simply mean the set of all such that for at least one .
The set and are obviously open in .
Let us prove [topology:openii]. If , then for all . As are all open, there exists a for every such that . Take and note that . We have for every and thus . Thus the intersection is open.
Let us prove [topology:openiii]. If , then for some . As is open then there exists a such that . But then and so the union is open.
The main thing to notice is the difference between items [topology:openii] and [topology:openiii]. Item [topology:openii] is not true for an arbitrary intersection, for example , which is not open.
The proof of the following analogous proposition for closed sets is left as an exercise.
[prop:topology:closed] Let be a metric space.
- [topology:closedi] and are closed in .
- [topology:closedii] If is an arbitrary collection of closed sets, then is also closed. That is, intersection of closed sets is closed.
- [topology:closediii] If are closed then is also closed. That is, finite union of closed sets is closed.
We have not yet shown that the open ball is open and the closed ball is closed. Let us show this fact now to justify the terminology.
[prop:topology:ballsopenclosed] Let be a metric space, , and . Then is open and is closed.
Let . Let . Of course . Now let . Then Therefore for every . So and is open.
The proof that is closed is left as an exercise.
Again be careful about what is the ambient metric space. As is an open ball in , this means that is an open set in . On the other hand is neither open nor closed in .
A useful way to think about an open set is a union of open balls. If is open, then for each , there is a (depending on of course) such that . Then .
The proof of the following proposition is left as an exercise. Note that there are other open and closed sets in .
[prop:topology:intervals:openclosed] Let be two real numbers. Then , , and are open in . Also , , and are closed in .
Connected sets
A nonempty metric space is connected if the only subsets that are both open and closed are and itself.
When we apply the term connected to a nonempty subset , we simply mean that with the subspace topology is connected.
In other words, a nonempty is connected if whenever we write where and and are open, then either or . So to test for disconnectedness, we need to find nonempty disjoint open sets and whose union is . For subsets, we state this idea as a proposition.
Let be a metric space. A nonempty set is not connected if and only if there exist open sets and in , such that , , , and
If is open in , then is open in in the subspace topology (with subspace metric). To see this, note that if , then as , we have . The proof follows by the above discussion.
The proof of the other direction follows by using to find and from two open disjoint subsets of .
Let be such that with and . Claim: is not connected. Proof: Notice
A set is connected if and only if it is an interval or a single point.
Suppose that is connected (so also nonempty). If is a single point then we are done. So suppose that and . If is such that , then is nonempty and is nonempty. The two sets are disjoint. As is connected, we must have they their union is not , so .
Suppose that is bounded, connected, but not a single point. Let and and note that . Suppose . As is the infimum, then there is an such that . Similarly there is a such that . We have shown above that , so . If , then as was the infimum, similarly if then . Therefore the only possibilities for are , , , .
The proof that an unbounded connected is an interval is left as an exercise.
On the other hand suppose that is an interval. Suppose that and are open subsets of , and are nonempty, and . We will show that and contain a common point, so they are not disjoint, and hence must be connected. Suppose that there is and . We can assume that . As is an interval . Let . If , then . If , then for any the ball contains points that are not in , and so as is open. Therefore, . As is open, for a small enough . As is the infimum of , there must exist some such that . Therefore . So and are not disjoint and hence is connected.
In many cases a ball is connected. But this is not necessarily true in every metric space. For a simplest example, take a two point space with the discrete metric. Then , which is not connected as and are open and disjoint.
Closure and boundary
Sometime we wish to take a set and throw in everything that we can approach from the set. This concept is called the closure.
Let be a metric space and . Then the closure of is the set That is, is the intersection of all closed sets that contain .
Let be a metric space and . The closure is closed. Furthermore if is closed then .
First, the closure is the intersection of closed sets, so it is closed. Second, if is closed, then take , hence the intersection of all closed sets containing must be equal to .
The closure of in is . Proof: Simply notice that if is closed and contains , then must contain and (why?). Thus . But is also closed. Therefore the closure .
Be careful to notice what ambient metric space you are working with. If , then the closure of in is . Proof: Similarly as above is closed in (why?). Any closed set that contains must contain 1 (why?). Therefore , and hence when working in .
Let us justify the statement that the closure is everything that we can “approach” from the set.
[prop:msclosureappr] Let be a metric space and . Then if and only if for every , .
Let us prove the two contrapositives. Let us show that if and only if there exists a such that .
First suppose that . We know is closed. Thus there is a such that . As we see that and hence .
On the other hand suppose that there is a such that . Then is a closed set and we have that , but . Thus as is the intersection of closed sets containing , we have .
We can also talk about what is in the interior of a set and what is on the boundary.
Let be a metric space and , then the interior of is the set The boundary of is the set
Suppose and . Then it is not hard to see that , , and .
Suppose with the discrete metric. Let , then and .
Let be a metric space and . Then is open and is closed.
Given we have such that . If , then as open balls are open, there is an such that , so is in . Therefore and so is open.
As is open, then is closed.
The boundary is the set of points that are close to both the set and its complement.
Let be a metric space and . Then if and only if for every , and are both nonempty.
If , then there is some such that as is closed. So contains no points of .
Now suppose that , then there exists a such that , but that means that contains no points of .
Finally suppose that . Let be arbitrary. By contains a point from . Also, if contained no points of , then would be in . Hence contains a points of as well.
We obtain the following immediate corollary about closures of and . We simply apply .
Let be a metric space and . Then .
Exercises
Prove . Hint: consider the complements of the sets and apply .
Finish the proof of by proving that is closed.
Prove .
Suppose that is a nonempty metric space with the discrete topology. Show that is connected if and only if it contains exactly one element.
Show that if is a connected unbounded set, then it is an (unbounded) interval.
Show that every open set can be written as a union of closed sets.
a) Show that is closed if and only if . b) Show that is open if and only if .
a) Show that is open if and only if . b) Suppose that is an open set and . Show that .
Let be a set and , be two metrics on . Suppose that there exists an and such that for all . Show that is open in if and only if is open in . That is, the topologies of and are the same.
Suppose that , is a collection of connected subsets of a metric space . Suppose that there exists an such that for all . Show that is connected.
Let be a connected set. a) Is connected? Prove or find a counterexample. b) Is connected? Prove or find a counterexample. Hint: Think of sets in .
The definition of open sets in the following exercise is usually called the subspace topology. You are asked to show that we obtain the same topology by considering the subspace metric.
[exercise:mssubspace] Suppose is a metric space and . Show that with the subspace metric on , a set is open (in ) whenever there exists an open set such that .
Let be a metric space. a) For any and , show . b) Is it always true that ? Prove or find a counterexample.
Let be a metric space and . Show that .
Contributors and Attributions
