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**johnsomeone** For A any subset of X, define the closure of A (denoted as cl(A)) by cl(A) = interesection of all closed sets containing A.

Note that cl(A) is always closed. Also, if A a subset of D, with D closed, then necessarily cl(A) is a subset of D. (Proofs are very easy)

Consider R with its usual topology. Let U = (p-e, p+e).

Note that [p-e, p+e] = the complement of { (-infinity, p-e) union (p+e, infinity) }.

Thus [p-e, p+e] is the complement of an open set, and hence is closed.

Since [p-e, p+e] is a closed set that obviously contains U (= (p-e, p+e)), it follows that U a subset of cl(U) a subset of [p-e, p+e].

But then there are only 4 possibilities for cl(U):

Possibility #1: cl(U) = A1 = (p-e, p+e) (= U).

Possibility #2: cl(U) = A2 = (p-e, p+e].

Possibility #3: cl(U) = A3 = [p-e, p+e).

Possibility #4: cl(U) = A4 = [p-e, p+e].

Show that possibilities #1-#3 cannot work, since A1, A2, A3 are not closed, and thus cl(U) = A4 = [p-e, p+e].

There are many ways to show that, but at this elementary description, lets just stick to the basic definitions.

Claim: A2 is not closed.

Proof:

ASSUME A2 = (p-e, p+e] is closed.

Then A2's complement, call it V, is open. V = (-infinity, p-e] union (p+e, infinity).

Since V is open in R, using that the basis for the topology is the open intervals, have that every x in V is contained in some open interval W that's entirely contained in V.

Let x = p-e in V. Then there exists an interval W containing x such that W is contained in V.

Let W = (x-a, x+b), where a and b are some positive real numbers such that x in W subset V.

Note that could have one or both of a & b being infinity, but if that were possible, then so would it be possible to choose a W' contained in W that also "worked" but that didn't have infinity at an interval endpoint.

Note that a & b both positive is required in order that x be in W.

But then W = (p-e-a, p-e+b) is a subset of V = (-infinity, p-e] union (p+e, infinity).

But that's clearly impossible, as will now show, and this contradiction will prove that the initial assumption was false, and thus that A2 is not closed.

Let D = min{2e, b}. Then D positive since both b and 2e are.

Let y = p-e+D/2. Will show that y in W, but y not in V, contradicting that W is a subset of V.

Note that D>0, so D/2>0, so from p-e < p-e+D/2(=y) < p-e+D/2+D/2 we know that p-e < y < p-e+D.

Then p-e-a < p-e < y < p-e+D <= p-e+b, proving that y in W = (p-e-a, p-e+b).

Also, p-e < y < p-e+D <= p-e+2e = p+e. Thus y in (p-e, p+e) which is a subset of A2. Thus y can't be in V, since V is the complement of A2.

Thus the assumption that A2 is closed has led to a contradiction.

Therefore A2 is not closed, and the claim is proven.

The exact same argument proves that A1 and A3 are not closed.

Therefore, cl(U) = A4 = [p-e,p+e], since A4 is the *only* closed set between U and the closed set A4.

Therefore cl(p-e,p+e) = [p-e,p+e].