Bastien Feb 1, 2021

# If A and B are open sets of a metric space X. Then(a) A ∩ B is open set(b) A ∪ B is open set(c) both (a) and (b)(d) none​

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Contributed by Jiří Lebl

Assistant Professor (Mathematics) at Oklahoma State University

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 (X,d) be a metric space, x∈X and δ>0 . Then define the open ball or simply ball of radius δ around x as

B(x,δ):={y∈X:d(x,y)<δ}.(8.2.1)

Similarly we define the closed ball as

C(x,δ):={y∈X:d(x,y)≤δ}.(8.2.2)

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 BX(x,δ):=B(x,δ) or CX(x,δ):=C(x,δ) .

Take the metric space R with the standard metric. For x∈R , and δ>0 we get

B(x,δ)=(x−δ,x+δ)andC(x,δ)=[x−δ,x+δ].(8.2.3)

Be careful when working on a subspace. Suppose we take the metric space [0,1] as a subspace of R . Then in [0,1] we get

B(0,\nicefrac12)=B[0,1](0,\nicefrac12)=[0,\nicefrac12).(8.2.4)

This is of course different from \(B_

(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 (X,d) be a metric space. A set V⊂X is open if for every x∈V , there exists a δ>0 such that B(x,δ)⊂V . See . A set E⊂X is closed if the complement Ec=X∖E is open. When the ambient space X is not clear from context we say V is open in X and E is closed in X .

If x∈V and V is open, then we say that V is an open neighborhood of x (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 [0,1)⊂R is neither open nor closed. First, every ball in R around 0 , (−δ,δ) contains negative numbers and hence is not contained in [0,1) and so [0,1) is not open. Second, every ball in R around 1 , (1−δ,1+δ) contains numbers strictly less than 1 and greater than 0 (e.g. 1−\nicefracδ2 as long as δ<2 ). Thus R∖[0,1) is not open, and so [0,1) is not closed.

[prop:topology:open] Let (X,d) be a metric space.

[topology:openi] ∅ and X are open in X .

[topology:openii] If V1,V2,…,Vk are open then

⋂j=1kVj(8.2.5)

is also open. That is, finite intersection of open sets is open.

[topology:openiii] If {Vλ}λ∈I is an arbitrary collection of open sets, then

⋃λ∈IVλ(8.2.6)

is also open. That is, union of open sets is open.

Note that the index set in [topology:openiii] is arbitrarily large. By ⋃λ∈IVλ we simply mean the set of all x such that x∈Vλ for at least one λ∈I .

The set X and ∅ are obviously open in X .

Let us prove [topology:openii]. If x∈⋂kj=1Vj , then x∈Vj for all j . As Vj are all open, there exists a δj>0 for every j such that B(x,δj)⊂Vj . Take δ:=min{δ1,…,δk} and note that δ>0 . We have B(x,δ)⊂B(x,δj)⊂Vj for every j and thus B(x,δ)⊂⋂kj=1Vj . Thus the intersection is open.

Let us prove [topology:openiii]. If x∈⋃λ∈IVλ , then x∈Vλ for some λ∈I . As Vλ is open then there exists a δ>0 such that B(x,δ)⊂Vλ . But then B(x,δ)⊂⋃λ∈IVλ 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 ⋂∞n=1(−\nicefrac1n,\nicefrac1n)={0} , which is not open.

The proof of the following analogous proposition for closed sets is left as an exercise.

[prop:topology:closed] Let (X,d) be a metric space.

[topology:closedi] ∅ and X are closed in X .

[topology:closedii] If {Eλ}λ∈I is an arbitrary collection of closed sets, then

⋂λ∈IEλ(8.2.7)

is also closed. That is, intersection of closed sets is closed.

[topology:closediii] If E1,E2,…,Ek are closed then

⋃j=1kEj(8.2.8)

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 (X,d) be a metric space, x∈X , and δ>0 . Then B(x,δ) is open and C(x,δ) is closed.

Let y∈B(x,δ) . Let α:=δ−d(x,y) . Of course α>0 . Now let z∈B(y,α) . Then

d(x,z)≤d(x,y)+d(y,z)<d(x,y)+α=d(x,y)+δ−d(x,y)=δ.(8.2.9)

Therefore z∈B(x,δ) for every z∈B(y,α) . So B(y,α)⊂B(x,δ) and B(x,δ) is open.

The proof that C(x,δ) is closed is left as an exercise.

Again be careful about what is the ambient metric space. As [0,\nicefrac12) is an open ball in [0,1] , this means that [0,\nicefrac12) is an open set in [0,1] . On the other hand [0,\nicefrac12) is neither open nor closed in R .

A useful way to think about an open set is a union of open balls. If U is open, then for each x∈U , there is a δx>0 (depending on x of course) such that B(x,δx)⊂U . Then U=⋃x∈UB(x,δx) .

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Ariuthris
Feb 1, 2021