Can anyone help me prove the following?

Suppose S is a set of numbers which contain exactly one element, then S is closed.

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- November 18th 2012, 11:03 PMduckyoneProofs -_____-
Can anyone help me prove the following?

**Suppose S is a set of numbers which contain exactly one element, then S is closed.** - November 18th 2012, 11:09 PMjakncokeRe: Proofs -_____-
I'm not sure i understand the question. Do you mean like S = { {1}, {2}, {3} } etc... ? or S = {1}.

- November 18th 2012, 11:11 PMduckyoneRe: Proofs -_____-
like S = {1}.

- November 18th 2012, 11:18 PMjakncokeRe: Proofs -_____-
- November 19th 2012, 08:21 AMHallsofIvyRe: Proofs -_____-
I can't make heads or tails out of what you are saying! "If S contains only one element then it is the empty set no?" No! The empty set has

**no**elements! You may be confusing "element" with "subset". And makes no sense because your S does not**contain**{x}, it contains only x.

Further, none of that has anything to do with whether such a set is closed or not.

duckyone, this is NOT in general true. It**is**true in a metric space so I will assume that is what you mean. There are a variety of ways, even in a metric space, of defining "closed" set and I don't know which you are using. One I like is this: we say that a point, p, is an "interior point" of set A if and only if there exist such that the neighborhood centered on p with radius is a subset of A. We say that p is an "**ex**terior" point of set A if and only if it is an interior point of the**complement**of A. Finally, we say that p is a "boundary" point of A if and only if it is**neither**an interior point or exterior point of A.

We then say that a set is "open" if and only if it contains**none**of its boundary points and that a set is "closed" if and only if it contains**all**of its boundary points. Here, Taking A= {x}, if , there is some fixed distance between x and y, d(x,y). Taking , it is clear that the neighborhood of y, with radius , does NOT contain x so that y is an exterior point. That is,**every**is an exterior point of A. That means that the set of boundary points of A is either empty or {x} itself. In either case, A contains its set of boundary points and so is closed.

If your course did not introduce "boundary points" other commonly used definition is that a set is closed if and only if its complement is open. And, of course, a set is open if and only if every point is an interior point. This case, the complement of A= {x} is the set all points**other**than x. And every such point is, as shown above, an interior point of the complement of A. - November 19th 2012, 02:13 PMDevenoRe: Proofs -_____-
let us suppose you are talking about a set of REAL numbers.

DEFINITION: a subset S of the real numbers R is open if and only if it is a union of (perhaps infinitely many) open intervals (a,b) = {x in R: a < x < b}.

DEFINITION: a subset S of R is closed if and only if R - S is open.

THEOREM: the set S = {x_{0}} is closed.

PROOF:

R - S = (-∞,x_{0}) U (x_{0},∞) = A U B.

consider the open intervals of the form: A_{k}= (x_{0}-k,x_{0}), where k is a positive integer.

then:

.

similarly, for B_{k}= (x_{0},x_{0}+k) we have:

.

so A and B are open, so A U B is also open (being an even BIGGER union of infinitely many open intervals), so R - S is open, so S is closed.

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note that HallsOfIvy's post is much more general, and subsumes mine (the real line is a special case of a metric space: for the interval (a,b) we may use the epsilon: ε = (b-a)/2 in which case-

(a,b) = ((a+b)/2 - ε,(a+b)/2 + ε) = {x in R: d(x,(a+b)/2) < ε}

the metric function d in this case is d(x,y) = |y - x|).

***************

in the most general formulation possible (point-set topology), it is NOT true that singleton sets need be closed. for example, we can declare the only subsets of a set X that are open are the null set and X itself. for {x} to be closed (where x is some element of X), we would need X - {x} to be open. if X contains more than just the single point {x},

then X - {x} is neither Ø nor X itself, and as such cannot be open. this unhappy state of affairs CAN happen (such a topology is called the "indiscrete topology" and can be imagined like so: X is a "tightly bound blob" whose elements are all "so tightly glued together we can't separate them").