1. ## Proofs -_____-

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

2. ## Re: Proofs -_____-

I'm not sure i understand the question. Do you mean like S = { {1}, {2}, {3} } etc... ? or S = {1}.

3. ## Re: Proofs -_____-

like S = {1}.

4. ## Re: Proofs -_____-

Originally Posted by duckyone
like S = {1}.
If S contains only one element then it is the empty set no? Since the empty set is a subset of every set.
if the element is the empty set. Then vacously $\displaystyle {\phi} + {\phi} = {\phi} \in S$

but even if they mean S = {x}, you merely say that $\displaystyle {x} \cup {x} = {x} \in S$ and $\displaystyle {x} \cup {\phi} = {x} \in S$

5. ## Re: Proofs -_____-

Originally Posted by jakncoke
If S contains only one element then it is the empty set no? Since the empty set is a subset of every set.
if the element is the empty set. Then vacously $\displaystyle {\phi} + {\phi} = {\phi} \in S$

but even if they mean S = {x}, you merely say that $\displaystyle {x} \cup {x} = {x} \in S$ and $\displaystyle {x} \cup {\phi} = {x} \in S$
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 $\displaystyle \{x\}\cup \{x\}= \{x\} \in S$ 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 $\displaystyle \delta> 0$ such that the neighborhood centered on p with radius $\displaystyle \delta$ 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 $\displaystyle y\ne x$, there is some fixed distance between x and y, d(x,y). Taking $\displaystyle \delta= d(x,y)/2$, it is clear that the neighborhood of y, with radius $\displaystyle \delta$, does NOT contain x so that y is an exterior point. That is, every $\displaystyle y\ne x$ 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.

6. ## Re: 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 = {x0} is closed.

PROOF:

R - S = (-∞,x0) U (x0,∞) = A U B.

consider the open intervals of the form: Ak = (x0-k,x0), where k is a positive integer.

then:

$\displaystyle \bigcup_{k \in \mathbb{Z}^+} A_k = A$.

similarly, for Bk = (x0,x0+k) we have:

$\displaystyle \bigcup_{k \in \mathbb{Z}^+} B_k = B$.

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.

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

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").