Cratylus
Junior Member
- Joined
- Aug 14, 2020
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Prove using the def. of closed set that S=[0,[MATH]\omega[/MATH]) is not a closed subset of X=[0,[MATH]\omega[/MATH]] with order topology(prob 4.a,pg 68)
4.1 Definition Let X. be a topological space and let x[MATH]\in[/MATH] X . A set N[MATH]\in[/MATH] X is a
neighborhood of x if there an open set U[MATH]\in[/MATH] X s.t x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N
Attempted Proof(via contradiction)
Let (X,T) be a topology.S being the set of countable ordinals, it is Hausdorff.(They many open sets contained in
S)
This implies that if x [MATH]\ne[/MATH]y, there are two open sets U and V, s.t x[MATH]\in[/MATH] U ,y[MATH]\in[/MATH] V
with U [MATH]\cap[/MATH] V=0. But X\S is open and x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N\S
and y[MATH]\in[/MATH]V[MATH]\subset[/MATH] N\S in S. with U [MATH]\cap[/MATH] V=0
But this is impossible . Thus S is not closed in X
I am using the book A First Course in Topology by Robert Conover
l can assume everythinh up to this point and info on ordinals.
Any help would be appreciated.
4.1 Definition Let X. be a topological space and let x[MATH]\in[/MATH] X . A set N[MATH]\in[/MATH] X is a
neighborhood of x if there an open set U[MATH]\in[/MATH] X s.t x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N
Attempted Proof(via contradiction)
Let (X,T) be a topology.S being the set of countable ordinals, it is Hausdorff.(They many open sets contained in
S)
This implies that if x [MATH]\ne[/MATH]y, there are two open sets U and V, s.t x[MATH]\in[/MATH] U ,y[MATH]\in[/MATH] V
with U [MATH]\cap[/MATH] V=0. But X\S is open and x[MATH]\in[/MATH] U[MATH]\subset[/MATH] N\S
and y[MATH]\in[/MATH]V[MATH]\subset[/MATH] N\S in S. with U [MATH]\cap[/MATH] V=0
But this is impossible . Thus S is not closed in X
I am using the book A First Course in Topology by Robert Conover
l can assume everythinh up to this point and info on ordinals.
Any help would be appreciated.