Can anyone help me prove the following?
Suppose S is a set of numbers which contain exactly one element, then S is closed.
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 "exterior" 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.
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|).
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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").