Infinite Union!
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Then which of the options describe the set
X
?
Clarification : All sets on the left are of the form [ 2 n + 1 1 , 2 n 1 ] for some integer n .
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3 solutions
Beautiful question +1 !!
Nice Question. :-)
start : from 1/infinite thats open , limit 1/infinite >> 0 and end in 1 thats close
Here X = ( 0 , 1 ] which is neigther open and nor closed in R . Hence this shows that, arbitrary i.e infinite, union of closed sets "need not" be closed.
Yes! This question is a 'dual' to your Infinite Intersection!
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Yes. :-)
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And we can construct an open set too!
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:-)
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@展豪 張 – Yes. but we must mention the space. Because a set may be closed or open with respect to metric space containing it and never on its own. For example, In R the set ( 0 , 1 ) is o p e n set while in space X = ( 0 , 1 ) U ( 2 , 5 ) the set ( 0 , 1 ) is "closed as well as open"!
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@Akash Patalwanshi
–
Oh thanks for reminding me this detail. If I don't remember it wrong,
S
is always both close and open on the topology on
S
.
To specify, I was talking on the whole real number set in my last comment. :)
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@展豪 張 – Which set S ? Whole space? Yes the whole space Which is generally denotes by X is both open as well as closed in topology on X . But N o t every set is closed as well as open on topology on X . If you prefer Discrete topology on any set X then yes! i.e In discrete space every set is both open as well as closed. :-)
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@Akash Patalwanshi – Yes I am referring to the whole space. By definition it is open. And since the empty set is also open, the whole space is closed. I am wondering is there an intuitive way to understand open and closed on a discrete topology? It's just a bit weird for me.
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@展豪 張 – The discrete metric just says that d ( x , x ) = 0
and
d ( x , y ) = 1 , x = y
So say open ball B has radius r . If r < 1 then the only point it contains is the point it's centred on. So any single point has a ball of some radius around it containing only that point. This is the same thing as , B ( x ) = { x } so we know that every singleton is open. And now we're actually done! Since now we know that any point x in a set A has a ball containing it, because we can always construct a ball that only contains x . Since all sets are open, their complements are open as well. This implies that all sets are also closed. I hope this helps you!
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@Akash Patalwanshi – Thank you so much! Your explanation is very clear. By the way, is there a geometric way to understand this metric?
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@展豪 張 – I think no :-(
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@Akash Patalwanshi
–
So sad.
Seems like geometric meaning of most metric (except the Euclidean metric) can destroy my brain.
I remember that
p
-adic metric has some interesting property: all triangles are isosceles, all points in a ball is a centre......
This discrete metric is more crazy: all triangles are equilateral (except those with repeated vertex), all points in a ball is a centre, ball is either a singleton or the whole space......
X = ( 0 , 1 ] , which is half open, half close.