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# question on open, closed sets watch

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.

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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