Prove (0,$\Omega$] is not a metric space

Cratylus

Junior Member
Joined
Aug 14, 2020
Messages
82
First l am not a student.l am 58 and have graduated in 1993
Second I use A First Course in Topology by Robert Conover.
I know Conover is not in the Math Geneology Project and there are some errors in the book https://www.mathgenealogy.org/

Lastly, my knowledge of ordinals is shaky.
l am trying to prove the following:
Prove ([MATH]0[/MATH],[MATH]\Omega[/MATH] ] ([MATH](0,\omega_1[/MATH] ] ) is not a metric space.

My attempt
Suppose to the contrary that ( [MATH]0[/MATH], [MATH]\Omega[/MATH] ] is a metric space
Let X= [MATH]\Omega[/MATH] then d(X,Y)=d(Y,X) is false X is limit ordinal and has no
successors
Contradiction
Thus ( [MATH]0[/MATH], [MATH]\Omega[/MATH] ] is not a metric space
Any help would be appreciated.
 
Last edited:
First l am not a student.l am 58 and have graduated in 1993
Second I use A First Course in Topology by Robert Conover.
I know Conover is not in the Math Geneology Project and there are some errors in the book https://www.mathgenealogy.org/

Lastly, my knowledge of ordinals is shaky.
l am trying to prove the following:
Prove ([MATH]0[/MATH],[MATH]\Omega[/MATH] ] ([MATH](0,\omega_1[/MATH] ] ) is not a metric space.

My attempt
Suppose to the contrary that ( [MATH]0[/MATH], [MATH]\Omega[/MATH] ] is a metric space
Let X= [MATH]\Omega[/MATH] then d(X,Y)=d(Y,X) is false X is limit ordinal and has no
successors
Contradiction
Thus ( [MATH]0[/MATH], [MATH]\Omega[/MATH] ] is not a metric space
Any help would be appreciated.
It is not a metric space because W,
the first uncountable ordinal,
does not have a countable local base.
 
Top