Point of post: This is a continuation of this post.
Point of post: In this post we complete the problems that appear at the end of Halmos, sections 32 and 33.
Point of post: In this post I will solve the, very few, problems in the twenty-eighth section of Halmos. This section is on the parity (sign) of a permutation.
Point of post: This is a continuation of this post.
Problem: Let . Let and
a) Show that and that equality holds iff
b) Show that and that equality holds iff is surjective
a) Let , then$ latex f(x)\in f\left(A_0\right)$ and so . Now, suppose that is injective and let clearly then . It follows by injectivity that (if this isn’t apparent, note that by definition means that for some . Now, by injectivity we see that and so the result becomes clear). Conversely, suppose that for every . Then, in particular we see that and so
from where injectivity follows.
b) Let , then for some . It follows then that .
Now, suppose that is surjective and let . Then, and so . Conversely, suppose that for every . Then, in particular we see that and surjectivity follows.
Problem: Let and let and for . Show that preserves inclusions, unions, intersections, and differences of sets:
Show that preserves inclusions and unions only:
g) ; show that equality holds iff is injective
h) ; show that equality holds iff is injective
a) Let , then and so and thus
b) See number three
c) See number three
d) Let then , or it follows that and (otherwise ). This is equivalent to saying that . Conversely, let , then and . Thus, and . It follows that or
e) Let then for some and thus (since ) we see that for some which is equivalent to saying that .
f) See number three
g) See number three for the first part. Clearly, it suffices to prove the second part for two sets. So, suppose that is injective and let then and so and and so and , or . The proposed equality follows.
Conversely, suppose that for every . We see then in particular the following line of reasoning
and clearly singletons are disjoint iff they’re single elements are not equal. Thus, condensing
and injectivity follows.
h) Let , then and . This, in particular says that for some and (otherwise . Thus, for some , or .
Now, suppose that is injective and let , then by injectivity and so and . Thus, and . So, . Conversely, if then we see that
but iff . It follows that and so .
Problem: Show that b), c), f), and g) of the last exercise hold for arbitrary unions and intersections
a) Let then . Thus, there exists some such that and so . Therefore, .
Conversely, if let , then for some . Therefore, so that . So finally, we may conclude that
c) We merely need note that
(where the first equality is gotten noticing that and using property d) from the previous problem). Then, recalling the last problem we can see that this is equal to
and so comparing the LHS of the first equality with the RHS of the last leads to the desired result.
f) Let . Then, for some . Thus, for some for some . Thus, and so .
Conversely, let . Then, for some . It follows that for some and so for some . Therefore,
g) Let . Then, for some . It follows that for some for every . Thus, for every and so
Problem: Let and
a) If , show that
b) If and are injective, show that is injective.
c) If is injective, what can you say about the injectivity of and ?
d) If and are surjective, prove that is surjective
e) If is surjective, what can you say about the surjectivity of and ?
a) Let , then but this clearly implies that . Conversely, if we see that and so . Therefore,
b) If then by the injectivity of we see that and so by the injectivity of it follows that .
c) If is injective then is injective. To see this, suppose not; then there exists such that but which contradicts ‘s injectivity.
d) We note that from where the conclusion follows.
e) It must be true that is surjective. To see this suppose not. Then,
which contradicts the surjectivity of
Problem: In general, let us denote the identity function on a set by . That is, define
Given we say that a function is a left inverse for if ; and we see that is a right inverse for if .
a) Show that if has a left inverse, is injective; and if has a right inverse, is surjective
b) Give an example of a function that has a left inverse but no right inverse
c) Give an example of a function which has a right inverse but no left inverse.
d) Can a function have more than one left inverse? More than one right inverse?
e) Show that if has both a left inverse and a right inverse , then is bijective and
a) If we see that and so is injective. Conversely, let be arbitrary, we know that and so is the required element which maps to under
b) The inclusion map . Clearly it has no right inverse or it’d be surjective. But, the mapping
surely satisfies the left inverse requirement since
c) Consider the function . Clearly this possesses no left inverse, otherwise it’d be injective. But, has the quality that and so is a suitable right inverse
d) Yes, in the first example we could have taken and the second example we could have had , yet both are left and right inverses respectively.
e) Clearly by a) and b) we see that is bijective. Furthermore, let be arbitrary (we can say since it’s surjective), then
where is furnished by ‘s injectivity. Thus, putting the two together gives .
Problem: Let . By restricting the domain and range of obtain from a bijective function .
Proof: Merely take and and so then the function is bijective and a restriction of (in fact, you could take the vaccuous function )
Problem: Amost all the laws of elementary arithmetic are consequences of the axioms defining a field. Prove, in particular, that if is a field, and if and belong to , then the following relations hold.
b) If then
g) If then either or
a) By axiom 3 (A3) we know that and by the commutativity described in A1 we conclude that
b) We see that if then which by associativity and commutativity says that which then implies that .
c) We use associativity and commutativity to rewrite our equations as
d) By commutativity of the multiplication it suffices to note that and thus and by associativity we arrive at .
e) We merely note that and thus .
f) We use e) to say that . Then, we notice that from where it follows that and thus and the conclusion follows.
g) Suppose that then since we see that which contradicts our choice of
a) Is the set of all positive integers a field?
b) What about the set of all integers?
c) Can the answers to both these question be changed by re-defining addition or multiplication (or both)?
a) No, we merely note that there is no additive identity for
b) No, there is no multiplicative identity for
c) Yes. But first before we justify let us prove a lemma (which is useful),
Lemma: Let be a field with . Then, given any set with there are operations for which is a field.
Proof: By virtue of their equal cardinalities there exists some bijection . Then, for define
We prove that with these operations is a field. We first note that and so they are legitimate binary operations. We now begin to show that all the field axioms are satisfied
1) Addition is commutative- This is clear since
2) Addition is associative- This is also clear since
which is equal to
which finally is equal to
3) There exists a zero element- Let be the zero element of then is clearly the zero element of . To see this we note that
for every .
4) Existence of inverse element- If we note that
which is the identity element of
5-8 are the analogous axioms for multiplication, which are (for the most part) the exact same as the above.
9) Distributivity- We note that
from where the rest is obvious.
This completes the lemma
Now, we may answer the question. Since is a field and the above lemma implies there exists addition and multiplications on and which make them into fields.
Problem: Let and let denote the integers .
a) Prove this is a field precisely when is prime
b) What is in ?
c) What is in ?
a) We appeal to the well-known fact that is solvable precisely when . From there we may immediately disqualify non-primes since the number of multiplicatively invertible elements of is and when is not a prime. When is a prime the only thing worth noting is that every non-zero element of has a multiplicative inverse. The actual work of showing the axioms hold is busy work, and I’ve done it before.
b) It’s clearly . Since
c) It’s . To see this we note that
Problem : Let be a field and define
show that either there is no such that or that if there is, the smallest such is prime
Proof: Assume that and . Now, suppose that where . We see then that
which upon expansion equals
which by associativity and grouping is equal to
which by concatenation of the equations yields
but since is a field it follows that or , either way the minimality of is violated.
a) Is a field?
b) What if are required to be integers?
a) This is a classic yet tedious exercise, I will not do it here.
b) No. For example, consider . Then, we have that
a) Doest the set of all polynomials with integer coefficients () form a field?
b) What about ?
b) No. I’ll let you figure these out (it’s really easy)
Let be the set of all ordered pairs of real numbers
a) Is a field if addition and multiplication are done coordinate wise?
b) If addition and multiplication are done as one multiplies complex numbers?
a) No. Consider that is a not the additive identity but it has no multiplicative inverse.
b) Yes, this is just field isomorphic to
Problem: Show that the normal formulation of continuity is equivalent to the open set version.
Proof: Suppose that are metric spaces and for every and every there exists some such that . Then, given an open set we have that . To see this let then and since is open by hypothesis there exists some open ball such that and thus by assumption of continuity there is some such that and so and thus is an interior point of .
Conversely, suppose that the preimage of an open set is always open and let and be given. Clearly is open and thus is open. So, since there exists some such that and so
Problem: Suppose that is continuous. If is a limit point of the subset of , is it necessarily true that is a limit point of ?
Proof: No. Consider with the suspace topology inherited from with the usual topology. Define
This is clearly continuous since and which are obviously open. But, notice that is a limit point for since given a neighborhood of we must have that there is some which contains it. But, is not a limit point for since that set has no limit points.
Problem: Let and denote a singlet set in the two topologies and respectively. Let be the identity function. Show that
a) is continuous if and only if is finer than
b) is a homeomorphism if and only if
a) Assume that is continuous then given we have that . Conversely, if is finer than we have that given that
b) If is a homeomorphism we see that both it and are continuous and so mimicking the last argument we see that and . Conversely, if then we now that
or equivalently that
which defines the homeomorphic property.
Problem: Given and show that the maps and given by and are topological embeddings.
Proof: Clearly and are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, then by definition of an ordered pair we must have that . Lastly, the inverse function is continuous since is the restriction of the projection to . The same is true for .
Problem: Show that with the usual subspace topology and .
Proof: Define and . These are easily both proven to be homeomorphisms.
Problem: Find a function which is continuous at precisely one point.
Suppose that is continuous at , then choosing sequences of rational and irrationals numbers respectively both converging to . We see by the limit formulation of metric space continuity that
And so if were to be continuous anywhere it would have to be at . To show that it is in fact continuous at we let be given then choosing we see that from where the conclusion follows since this implies that .
a) Suppose that is “continuous from the right”, that is, for each . Show that is continuous when considered as a function from to .
b) Can you conjecture what kind of functions are continuous when considered as maps as . As maps from to ?
a) Note that by the assumption that we know that for every there exists some such that implies that . So, let be open and let . Then, and since is open we see that there is some such that . But, by assumption there exists some such that . But, and thus and thus and so is an interior point for from where it follows that is open and thus is continuous.
b) I’m not too sure, and not too concerned right now. My initial impression is that if is continuous then is open which should be hard to do. Etc.
Problem: Let be an ordered set in the order topology. Let be continuous.
a) Show that the set is closed in
b) Let . Show that is continuous.
a) Let then . Suppose first that there is no and consider
This is clearly open in by the continuity of and is contained in it. Now, to show that let then which with simplification gives the important part that and so but since there is no such that this implies that . Similar analysis shows that and since there is no as was mentioned above this implies that . Thus, and thus .
Now, suppose that there is some such that then letting we once again see that is open and . Furthermore, a quick check shows that if that and so and and so and so so that . The conclusion follows
b) Let and . As was shown in a) both are closed and thus define
Notice that since are both assumed continuous and that we may conclude by the gluing lemma that is in fact continuous. But, it is fairly easy to see that
Problem: Let be a collection of subset of ; let . Let and suppose that is continuous for each
a) Show that if the collection is finite each set is closed, then is continuous.
b) Find an example where the collection is countable and each is closed but is not continuous.
c) An indexed family of sets is said to be locally finite if each point of has a neighborhood that intersects only finitely many elements of . Show that if the family is locally finite and each is closed then is continuous.$
a) This follows since if is closed then it is relatively easy to check that but since each is continuous we see that is closed in . But, since each is closed in it follows that each is closed in . Thus, is the finite union of closed sets in , and thus closed.
b) Give the subspace topology inherited from with the usual topology and consider with
Clearly each i
Lemma: Let be any topological space and be a locally finite collection of subsets of . Then,
Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let since the collection of sets is locally finite there exists some neighborhood of such that it intersects only finitely many, say , elements of the collection. So, suppose that then is a neighborhood of which does not intersect contradicting the assumption it is in the closure of that set.
Now, once again we let be closed and note that and each is closed in and since is closed in we see that is closed in . So, noting that it is evident from the assumption that is locally finite in that so is and thus (for notational convenience) letting the above lemma implies that
From where it follows that the preimage of a closed set under is closed. The conclusion follows.
Problem: Let and be continuous functions. Let us define a map by the equation . Show that is continuous.
Proof: This follows from noting the two projections of are and . But, both of these are continuous since . To see this we note that if and only if which is true if and only if or in other words . Using this we note that the preimage an open set in will be the product of open sets by the continuity of . It clearly follows both projections, and thus the function itself are continuous.
Problem: Let . We say that is continuous in eahc variable separately if for each , the map is continuous and for each the map is continuous. Show that if is continuous then is continuous in each variable separately.
Proof: If is continuous then clearly it is continuous in each variable since if we denote by the mapping we see that where but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable.
Problem: Let be given by
a) Show that is continuous in each variable separately.
b) Compute .
c) Show that is not continuous
a) Clearly both and are continuous for since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at since it is trivial to check that $
c) This clearly proves that is not continuous with is not continuous since if is the diangonal we have that
and so in particular
Problem: Let ; let be continuous and let be Hausdorff. Prove that if may be extended to a continuous function , then is uniquely determined by .
Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways
Way 1(fun way!):
Lemma: Let be any topological space and a Hausdorff space. Suppose that are continuous and define . Then, is closed in
Proof: Note that is clearly continuous since and . It is trivial then to check that and since is Hausdorff we have that is closed and the conclusion follows.
From this we note that if agree on such that we have that
From where it follows that and so . So, thinking of as a subspace of we see that and thus clearly is dense in . So, the conclusion readily follows by noting that if are two continuous extensions then by definition .
Way 2(unfun way): Let be two extensions of and suppose there is some . Clearly and thus is a limit point of . So, by assumption and so using the Hausdorffness of we may find disjoint neighborhoods of them respectively. Thus, are neighborhoods of in . Thus, is a neighborhood of . But, clearly there can be no otherwise . It follows that is a neighborhood of disjoint from which contradicts the density of in . The conclusion follow.
Problem: Show that if is a subspace of , and is a subspace of , the the topology it inherits as a subspace of is the same as it inherits as a subspace of .
Proof: Let and be the induced topologies on as a subspace of respectively. Let then for some open set in . But, and since is open in it follows that . Conversely, suppose that then for some open set in . But, since is a subspace of we know that for some open set in . Thus, and thus . The conclusion follows.
Problem: If are topologies on and is strictly finer than , what can you say about the corresponding subspace topologies on the subset of ?
Proof: Let and be the subspace topologies on inherited from respectively. Let then for some and thus . Thus, . The inclusion need not be strict. For example consider the set with the two topologies
Now, consider then
And thus . The inclusion will be strict if intersects every element of .
Problem: Consider the set as a subspace of . Which of the following sets are open in ?
We only bother to state which are open in since the results should be clear for to any student in pre-algebra.
is open sine .
is open since
is not open. To see this suppose that it was then is closed in , but since is closed in this would imply that was closed in .
is not open since if it were it’s complement would be closed in and since is closed in it follows is closed in .
: This is open since the last of which is open.
Problem: A map is said to be an open map if for every open set of , the set is open in . Show that and are open maps.
Proof: Assuming that has the product topology we know that given an open set we know that where are open in respectively. Thus,
Which are open in respectively.
Problem: Let and denote a simle set in the topologies respectively; let and denote a singe set in the topologies respectively. Assume that these sets are non-empty. Prove that if and then the product topology is finer than the product topology on
Proof:It suffices to check that each basic element in has a basic element in contained within it which contains and arbitrary point. So, let be open in and let $. Then, we see by the previous problem that and and thus since the topologies on are finer than those on there are basic open sets in respectively such that and . Thus, and since is basic open in the conclusion follows.
Problem: Show that is an open base for
Proof: This follows from the fact that is an open base for and has the product topology.
Problem: Let be an ordered set. If is a proper subset of that is convex in does it follow that is an interval or a ray in ?
Proof: No. Consider with the usual ordering and the set . This is clearly convex since if then for any we have that and thus or in other words . But, it is not an interval or a ray. Clearly if it were either it must be a ray, so assume that or . In both cases we must have that and thus by the density of the rationals in we may find such that or in other words that and thus .
Problem: If is a straight line in the plane describe the topology inherets as a subspace of and as a subspace of . In each case it is a familiar topology.
Proof: Clearly as was shown since and are open bases for and we know that is an open base for . Now, if you’ll recall the subspace topology on can be described by . So, we break this into three cases based on what kind of line is.
So, firstly suppose that is a vertical line then for some . So, we now claim . So, let then we have that
Conversely, let . Then,
Either way we see that .
A similar analysis shows that if that
We now break this into the last case where for some . So, once again we know that the subspace topology on will be that generated by . We now get slightly more informal. The key point is that since none of the corners of the rectangle are included the line, if it intersects, it intersects an interval like region of the line. Thus, a quick breakdown into the cases where it intersects the solid line and when it doesn’t should quickly convince you that the topology on will end up being the lower-limit topology.
Now, for the case with we have quite a different scenario. For example consider the anti-diagonal and consider for each fixed the basic open set clearly then from where it follows that is in fact a discrete space. In fact for any line that is non-horizontal and non-vertical we may find for each point on the line a basic open set whose intersection with the line is that single point. In other words the lines in inherit the discrete topology as subspaces. For horizontal and vertical lines they still have the lower limit topology.
Problem: Show that the dictionary order topology on the set is the same as the product topology on where denotes with the discrete topology. Compare this topology to the usual topology on
Proof: Let and let be a basic open neighborhood of it. Then, consider the set clearly this is open in with the lexicographic ordering and if we must have that which implies that and thus . Thus, noting finishes that portion of the argument. Conversely, let and be some basic open neighborhood containing it. If then and we’re done. So, assume that . If then consider . Clearly . Now, to see that we note that and thus and since we automatically have that .
Now, if then choosing and applying similar techniques works.
Lastly, if then choosing automatically works since and if we have that and . Regardless, we’ve found a basic open set in which contains and which is contained in .
It follows that the each topology is finer than the others, or that the topologies are equal.
The order topology on is finer than that of the usual topology on . Just note that if and is an open ball centered at it that is a basic open set in containing and which is contained inside the ball. The converse is not true. For example consider the origin and the basic open neighborhood . Any open ball centered at zero will contain points not on that line from where the conclusion follows.
Problem: Let . Compare the product topology on , the dictionary order topology on , and the topology inherits as a subspace of in the dictionary order topology.
Proof: Let denote the topologies in the order they were mentioned. We first claim that . To see this let be arbitrary and note that generates . Then, if we have that . To see why the inclusion’s strict we note as in the last example that for example and any open square must intersect points of not in that interval.
Clearly then as was shown in the book.
This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being.
Problem: Let be a topological space, let be a subset of . Suppose that for each there is an open set containing such that . Prove that is open.
Proof: We claim that but this is obvious since for each we have that . Conversely, since each we have that the union of all of them is contained in , namely . Thus, is the union of open sets and thus open.
Problem: Compare the nine topologies on given in example 1.
Solution: This is simple.
Problem: Show that given a set if we denote to be cocountable topology (a set is open if it’s complement is countable or the full space) that is a topological space. Is it still a topological space if we let ?
Proof: Clearly for the first part . Now, if is a collection of open sets then we note that is finite and thus for any . Thus, is finite and thus in . Now, if we have that and thus is the finite union of finite sets and thus finite, so .
If we redefine the topology as described it is not necessarily a topology. For example, give that topology and note that is a collection of elements of but whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology.
a) If is a family of topologies on , show that is a topology on . Is ?
b) Let be a family of topologies on . Show that there is unique topology on containing all the collections , and a unique largest topology contained in all of the .
c) If , let and . Find the smallest topology containing and the larges topology contained in .
a) Let and let be arbitrary. Then, by assumption we have that for every but since this was assumed to be a topology we have that for every that and thus . Now, if we must have that for every and thus for every and thus . Thus, noting that for every we must have that the conclusion follows.
No, the union of two topologies needn’t be a topology. Let be defined as in part c and note that but
b) This follows immediately from part a. For the first part let (where is the set of all topologies) and let , this clearly satisfies the conditions. For the second one merely take
c) We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact . But this is just computation and we leave it to the reader. For the second part just intersect the two topologies.
Problem: Show that if is a base for a topology on , then the topology generated by equals the intersection of all topologies on which contain . Prove the same if is a subbase
Proof: Let be the intersection of all topologies on which contain and the topology generated by . Clearly since is itself a topology on containing . Conversely, let we show that where is any topology on containing . But, this is obvious since for some and thus is the union of open sets in and thus in . The conclusion follows.
Next, let be above except now is a subbase. For the same reasons as above we have that . Conversely, for any topology containing we have that if then where each is the finite union of elements of . But, by construction it follows that each is open (it is the finite intersection of open sets in ) and thus is the union of open sets in and thus .
Problem: Show that the topologies on and the Sorgenfrey line aren’t comparable
Proof: See the last part of the next problem
Problem: Consider the following topologies on:
For each determine which of the others contain it.
:Clearly we have that since the defining open base for is contained entirely in .
: But, since but and thus not in . Now, to prove the inclusion indicated we know that for each open set in the cofinite topology we have that and so if we assume WLOG that then
And thus is open in the usual topology.
: But, . To see this it suffices to show that is open in since this is a base for . But, to see this we must merely note that .
: Lastly, . To see this we must merely note that .
For the result is obvious except possibly how it relates to . But, in fact they aren’t comparable. To see this we first show that is not open in . To see this we show it can’t be written as the union of sets of the form and but it clearly suffices to do this for the latter sets. Now, to see that can’t be written as the union of sets of the form we recall from basic real number topology that is not open ( is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on . Also, consider .
Problem: Show that the countable collection is a base for the usual topology on
Proof: This follows from the density of . It suffices to show that given any and any that there is some element such that . But, from basic analysis we know there is some rational number such that and similarly there is some such that . Thus, . The conclusion follows. .
Problem: Show that the collection generates a different topology from the one on (the lower limit topology)$.
Proof: Clearly is open in but we show that it can’t be written as the union of elements of . So, suppose that where . Then, there exists some such that . Now, since we must have that but and thus and thus . Contradiction.
Problem: Prove that ever closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is a set which is countable.
Proof: It is easy to see that the above proof (proof 27.) holds in any space which is second countable, in particular for separable metric spaces. Thus, if is closed we surely have that . Thus, but since and we have then that which is the union of a perfect and countable set respectively.
Problem: Prove that every open set may be written as the disjoint union of countably many open intervals.
Proof: We need to prove a quick lemma
Lemma: Let be a topological space and let be a class of connected subspace of such that then is a connected subspace of .
Proof: Suppose that is a separation of . We may assume WLOG that for some . So, now we see that otherwise would be non-empty disjoint subsets of whose union is contradicting that is connected. Thus, it easily follows that for any we have that so that and thus by a similar reasoning we see that . Thus, since was arbitrary it follows that contradicting that .
So, now for each define
And let and finally we prove that
is a countable class of disjoint open intervals whose union is . The fact that each is an open interval follows since it is the union of open connected spaces with non-empty intersection (each element of contains ) it is also an open connected subspace of (note that each element of is open in but since is open it is also open in . But, it was proven in the book the only connected subspace of are intervals and thus is an interval for each .
Now, to see that they are disjoint we show that if then from where the conclusion will follow. So, to see this we first note that if is non-empty then is an open connected subspace of containing both and and thus and but this implies that and respectively from where the conclusion follows.
Now, to see that is countable we notice that for each we have that and so if we let be a fixed but arbitrary then
is an injection since the elements of are pairwise disjoint. The fact that is countable follows immediately.
Thus, is a countable collection of open intervals and
Thus, the conclusion follows.
Problem: Prove that if where each is a closed subset of then at least one has non-empty interior.
Lemma: Let be a complete metric space and a descending sequence of non-empty closed subsets of such that . Then, contains one point.
Proof: Clearly does not contain more than one point since . So, now for each choose some and let . Now, if were somehow finite a similar argument to that used in the first case of problem 26. is applicable, so assume that is infinite. But, since it is evident that is a Cauchy sequence and thus by assumption it converges to some point . We claim that . To see this we note similarly to problem 26. that since is infinite it is easy to see that is a limit point of and so if for some then is a neighborhood of containing only finitely many points of which clearly contradicts that it is a limit point. The conclusion follows.
So, now suppose each had empty interior (i.e. nowhere dense) . Then, since is open and is nowhere dense there exists an open ball of radius less than one such that . Let be the concentric closed ball of whose radius is half that of . Since is nowhere dense contains an open ball of radius less than one-half which is disjoint from . Let be the concentric closed ball of whose radius is one-half that of . Since is nowhere dense we have that contains an open ball of radius less than one-fourth which is disjoint from . Let be the concentric closed ball of whose radius is half that of . Continuing in this we get a descending sequence of non-empty closed subsets for which . Thus, by our lemma we have that there is some . This point is clearly not in any of the ‘s from where the conclusion follows.