# question on open, closed setsWatch

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#1
Hello there,
I understand the definition of open and closed sets in terms of metric spaces, but I find them hard to apply to specific questions!!

The definition of an open set is as follows:
"Let (X,d) be a metric space. Then a subset (Y,d) of (X,d) is open if for all x E (Y,d) there exists an epsilon > 0 such that the epsilon ball about the point x is a subset of (Y,d) . "

And " a subset (Y,d) is closed if (X,d)\(Y,d) is open."

As I say, I find these defintions difficult to apply when trying to determine whether sets are open or closed. For example, consider the following questions:

{x E Q | x^2 <= 3} in the space Q with the Euclidean Metric. (Q is the rationals here)

{(x,y) E R^2 | 2<=x , 0<=y<=1/x } in the space R^2 with the Euclidean metric.

So I'm not sure how to show these sets are open or closed from the definition!!

Any help would be much appreciated.

0
13 years ago
#2
If you understand open/closed subsets in R then it's easy to understand the open/closed subsets of A, where A is a subset of R.

Let S be a subset of A, which in turn is a subset of R.

S is open (resp. closed) in A if and only if there exist exists an open (resp. closed) subset T of R such that

S = A n T.

(Original post by beast)

S = {x E Q | x^2 <= 3} in the space Q with the Euclidean Metric. (Q is the rationals here)
T = (-rt(3),rt(3)) is open in R;
S = T n Q;
so S is open in Q.

U = [-rt(3),rt(3)] is closed in R;
S = U n Q;
so S is closed in Q.

S is open and closed in Q.

(Original post by beast)
V = {(x,y) E R^2 | 2<=x , 0<=y<=1/x } in the space R^2 with the Euclidean metric.
This is done in R^2, so no need to worry about subspace topologies like in the previous example. You can think about putting open discs around points of V and R^2 - V.

Sketch V first that should help. Then note:

any disc around (2,0) (or indeed any point of the boundary of V) will contain some of the complement of V; that is, (2,0) is not an interior point of V and so V is not open.

given any point in the complement of V, the a small enough disc can be placed around that point and still in the complement. The complement is open and so V is closed. Another way of looking at it is that V contains its limit points (includes its boundary).

So V is closed but not open in R^2
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