# Thread: closed and open points!!

1. ## closed and open points!!

1) Let S and T be subsets of R. Find a counter example for each of the following.

a) If P is the set of all isolated points of S, then P is a closed set
b) If S is closed, then cl (int S) = S
c) if S is open, then int (cl S) = S
d) bd (cl S) = bd S
e) bd (bd S) = bd S
f) bd (S U T) = (bd S) U (bd T)
g) bd ( S-intersect-T) = (bd S)-intersect-( bd T)

2) Prove:

a) S is closed iff S = cl S
b) cl S = S U bd S

2. Can you give the definition of:
Closed Set.
Closure of a set.
Open set
Interior of a set?
Boundary of a set.
Isolated Point.

Can you tell us what tools you have:
To prove a set is closed.
To prove a set is open
To prove a point is an interior point.
To prove a point is a boundary point.
To prove a point is an isolated point.

Definitions vary so much from textbook to textbook.

3. closed: Let S be a subset of R. If bd S subset of S, then S is said to be closed.

Open: If bd S is a subset of R\S, then S is open

Let S be a subset of R. A point x in R is an interior point of S if there exists a neighborhood N of x such that N is an element of S.

If for every neighborhood N of x, N-intersects-S is not equal to the empty set and N-intersects-(R\S) is not equal to empty set, then x is called a boundary point of S

Let S be a subet of R. Then the closure of S, denoted cl S, is defined by
cl S = S U S'

If x is and element of S and x is not an element of S', then x is called an isolated point of S

4. A set S is open iff S = int S. Equivalently, S is open iff every point in S is an interior point
of S.
A set S is closed iff its complement R\S is open
Problem one here is the hint I have:
parts a-e, think about the setes {5}, (3,5) U (5,7), Q, {1/n: n is an
element of N}
To prove B, you have to show that cl S is a subset S U bd S and that S U bd S is a subset cl S.
To show that cl S is a subset S U bd S, take an arbitrary point x is an element cl S
and deduce that x is an element of S U bd S – one way to do this is the following:
since by definition cl S = S U S',
x is an element of cl S implies that either x is an element S
or x is an element of S' \ S consider each of these two possibilities
separately and show that in each case you can conclude that
x is an element of S U bd S by using
only the definitions of bd S and S'. Note that the two possibilities x is and element of S
and x is an element of S'\S
are mutually exclusive, i.e., cannot occur simultaneously.

5. I have taught courses in topology for over thirty years. That is the strangest set of definitions I have ever seen. Of course they are, in fact, equivalent to other “standard” definitions.

Let’s see how to use them to prove “a) If P is the set of all isolated points of S, then P is a closed set.”
If x is an isolated point of S then because x is not in S’, there is a neighborhood N of x that contains no other point of S. That means that x is a boundary point of P. So bd(P)=P, which proves that P is closed because its boundary is a subset of P.

6. Right, I am trying to figure out the rest of them, I am just drawing a blank, not sure how to do them...This is just for review as I have a test coming friday.

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