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Closed and open sets

  • Thread starter yifli
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  • #1
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Assume X is a metric space, then X and the empty set are both closed and open,
am I correct?
 

Answers and Replies

  • #2
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Hi yifli, :smile:

That is indeed correct!
 
  • #3
HallsofIvy
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In general point set topology the fact that the topology (the collection of open sets) contains both the empty set and the entire space is part of the definition.
 
  • #4
How do we define closed sets?

A closed set is one that includes all its boundary points,is this definition right?

which of the following are closed sets?

a){(x,y): x^2+y^2 >=4}
b){(x,y):x^2+y^2<=4}
c){(x,y):x^2+y^2=4}
 
  • #5
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How do we define closed sets?

A closed set is one that includes all its boundary points,is this definition right?
Correct.

which of the following are closed sets?

a){(x,y): x^2+y^2 >=4}
b){(x,y):x^2+y^2<=4}
c){(x,y):x^2+y^2=4}
So, what do you think??
 
  • #6
a) {(x,y):x^2+y^2>=4}

x^2+y^2= 4 is equation of a circle in 2-dimensional space, with radius 2 and centre at (0,0)

the set given above shall include all points on the circle and beyond (extending to infinity);if i take a boundary point say,

(2,0) from this set,then this should be part of the set and it is.So can we say it is a closed set?

b) {(x,y):x^2+y^2<=4}

the set given above shall include all points on the circle and inside;if i take a boundary point say ,again, (0,2) from this set,then this should be part of the set and it is.So can we say it is a closed set?


c){(x,y):x^2+y^2=4}

the set given above shall include only the points on the circle;there isn't a boundary as such

what do we do in this case?
 
  • #7
A set is open if it is a neighbourhood of each of its points.Is this similar to saying that a set is open

if it contains all of its interior points?If yes,then can we proceed in the following way:

the set {(x,y): x^2+y^2>4} is in R^2

If i take a point ,say (q,s) which lies very close to the boundary point (0,2),then a

circle(neighbourhood of the point,subset of the original set) with this point as centre should also

lie entirely within the set.

But i couldn't get one thing -size of circle with (q,s) as centre is determined by what?In other words, what radius should we choose of this circle, since it will depend upon the size of this circle whether it is lying entirely in the bigger circle,that is, our original set or not( am i going in the right direction?)

If this proves our set to be open then according to definiton of closed set-complement of an open set is closed
{(x,y): x^2+y^2<=4} is complement of {(x,y):x^2+y^2>4},so if {(x,y):x^2+y^2>4} is an open set (according to procedure followed above),then {(x,y): x^2+y^2<=4} is a closed set.

I don't know whether my reasoning is right.Someone please guide.
 
  • #8
An open set is a set which is neighbourhood of all its points

a) {(x,y):x^2+y^2>4} is an open set because it is neighbourhood of all its points

A closed set is a set if its complement is an open set

b){(x,y):x^2+y^2<=4}is a closed set because its complement {(x,y):x^2+y^2>4} is an open set.

c){(x,y): x^2+y^2<4} is an open set because it is neighbourhood of all its points.

d){(x,y):x^2+y^2>=4} is a closed set because its complement {(x,y): x^2+y^2<4} is an open set

e){(x,y): x^2+y^2=4} is a closed set because complement of this set
includes points represented by the sets {(x,y):x^2+y^2>4} and {(x,y):x^2+y^2<4} is an open set.


Is my reasoning right?
 
  • #9
for set : {(x,y): x^2+y^2=4} ,the complement is given by union of two open sets
{{(x,y):x^2+y^2>4}union {(x,y):x^2+y^2<4}} which is an open set as per the definiton hence, {(x,y): x^2+y^2=4} is a closed set!
 

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