GRE 9367 #62: Prove $X=[0,1]$ in lower limit topology ($[a,b)$) is not compact, is Hausdorff and is...












1












$begingroup$


GRE9367 #62




enter image description here




Ian Coley's solution:




enter image description here




Sean Sovine's solution:




enter image description here






  1. Prove $X$ is not compact.


My first proof was similar to Ian Coley's, but I came up with another proof:




If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $mathbb R$. This implies that the topologies are equal by (*), a contradiction.




Did I go wrong somewhere?




  1. Prove $X$ is Hausdorff.


My proof is similar to Sean Sovine's. For Ian Coley's proof, is my understanding right?




If there exists the required open sets in standard topology, then we can choose the same sets as the required open sets in the lower limit topology.





  1. Prove $X$ is disconnected.



My proof is the same as Ian Coley's. Is Ian Coley's proof right?





(*) Munkres Exer26.1 (dbfin pf)




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general-topology compactness connectedness gre-exam separation-axioms







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  • $begingroup$
    @5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:03


















1












$begingroup$


GRE9367 #62




enter image description here




Ian Coley's solution:




enter image description here




Sean Sovine's solution:




enter image description here






  1. Prove $X$ is not compact.


My first proof was similar to Ian Coley's, but I came up with another proof:




If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $mathbb R$. This implies that the topologies are equal by (*), a contradiction.




Did I go wrong somewhere?




  1. Prove $X$ is Hausdorff.


My proof is similar to Sean Sovine's. For Ian Coley's proof, is my understanding right?




If there exists the required open sets in standard topology, then we can choose the same sets as the required open sets in the lower limit topology.





  1. Prove $X$ is disconnected.



My proof is the same as Ian Coley's. Is Ian Coley's proof right?





(*) Munkres Exer26.1 (dbfin pf)




enter image description here







general-topology compactness connectedness gre-exam separation-axioms







share|cite|improve this question











$endgroup$












  • $begingroup$
    @5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:03
















1












1








1


1



$begingroup$


GRE9367 #62




enter image description here




Ian Coley's solution:




enter image description here




Sean Sovine's solution:




enter image description here






  1. Prove $X$ is not compact.


My first proof was similar to Ian Coley's, but I came up with another proof:




If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $mathbb R$. This implies that the topologies are equal by (*), a contradiction.




Did I go wrong somewhere?




  1. Prove $X$ is Hausdorff.


My proof is similar to Sean Sovine's. For Ian Coley's proof, is my understanding right?




If there exists the required open sets in standard topology, then we can choose the same sets as the required open sets in the lower limit topology.





  1. Prove $X$ is disconnected.



My proof is the same as Ian Coley's. Is Ian Coley's proof right?





(*) Munkres Exer26.1 (dbfin pf)




enter image description here







general-topology compactness connectedness gre-exam separation-axioms







share|cite|improve this question











$endgroup$




GRE9367 #62




enter image description here




Ian Coley's solution:




enter image description here




Sean Sovine's solution:




enter image description here






  1. Prove $X$ is not compact.


My first proof was similar to Ian Coley's, but I came up with another proof:




If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $mathbb R$. This implies that the topologies are equal by (*), a contradiction.




Did I go wrong somewhere?




  1. Prove $X$ is Hausdorff.


My proof is similar to Sean Sovine's. For Ian Coley's proof, is my understanding right?




If there exists the required open sets in standard topology, then we can choose the same sets as the required open sets in the lower limit topology.





  1. Prove $X$ is disconnected.



My proof is the same as Ian Coley's. Is Ian Coley's proof right?





(*) Munkres Exer26.1 (dbfin pf)




enter image description here







general-topology compactness connectedness gre-exam separation-axioms




general-topology compactness connectedness gre-exam separation-axioms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 12 '18 at 17:10









José Carlos Santos

159k22126231




159k22126231










asked Oct 25 '18 at 10:57









BCLCBCLC

1




1












  • $begingroup$
    @5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:03




















  • $begingroup$
    @5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:03


















$begingroup$
@5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
$endgroup$
– BCLC
Oct 25 '18 at 11:03






$begingroup$
@5xum Munkres Exer26.1 or Compact Hausdorff Topology is Minimal Hausdorff or something. I'll update my question.
$endgroup$
– BCLC
Oct 25 '18 at 11:03












2 Answers
2






active

oldest

votes


















1












$begingroup$



  1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.

  2. Yes, it is fine.

  3. It is fine too.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:05






  • 1




    $begingroup$
    Yes, that part was missing, as I wrote in my answer.
    $endgroup$
    – José Carlos Santos
    Oct 25 '18 at 11:06










  • $begingroup$
    Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:07



















1












$begingroup$

$[0,1]$ is not connected, e.g. because $[0,frac12)$ is a non-trivial closed-and-open subset of it.



Not compact as ${1-frac1n: n=2,3,4,5,ldots}$ is an infinite subset without limit point in it.



Hausdorff because as you state its topology includes the usual, Hausdorff, one.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 12:08













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$



  1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.

  2. Yes, it is fine.

  3. It is fine too.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:05






  • 1




    $begingroup$
    Yes, that part was missing, as I wrote in my answer.
    $endgroup$
    – José Carlos Santos
    Oct 25 '18 at 11:06










  • $begingroup$
    Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:07
















1












$begingroup$



  1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.

  2. Yes, it is fine.

  3. It is fine too.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:05






  • 1




    $begingroup$
    Yes, that part was missing, as I wrote in my answer.
    $endgroup$
    – José Carlos Santos
    Oct 25 '18 at 11:06










  • $begingroup$
    Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:07














1












1








1





$begingroup$



  1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.

  2. Yes, it is fine.

  3. It is fine too.






share|cite|improve this answer









$endgroup$





  1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.

  2. Yes, it is fine.

  3. It is fine too.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 25 '18 at 11:03









José Carlos SantosJosé Carlos Santos

159k22126231




159k22126231












  • $begingroup$
    Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:05






  • 1




    $begingroup$
    Yes, that part was missing, as I wrote in my answer.
    $endgroup$
    – José Carlos Santos
    Oct 25 '18 at 11:06










  • $begingroup$
    Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:07


















  • $begingroup$
    Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:05






  • 1




    $begingroup$
    Yes, that part was missing, as I wrote in my answer.
    $endgroup$
    – José Carlos Santos
    Oct 25 '18 at 11:06










  • $begingroup$
    Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 11:07
















$begingroup$
Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
$endgroup$
– BCLC
Oct 25 '18 at 11:05




$begingroup$
Munkres Exer26.1? I updated my question. Thanks José Carlos Santos!
$endgroup$
– BCLC
Oct 25 '18 at 11:05




1




1




$begingroup$
Yes, that part was missing, as I wrote in my answer.
$endgroup$
– José Carlos Santos
Oct 25 '18 at 11:06




$begingroup$
Yes, that part was missing, as I wrote in my answer.
$endgroup$
– José Carlos Santos
Oct 25 '18 at 11:06












$begingroup$
Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
$endgroup$
– BCLC
Oct 25 '18 at 11:07




$begingroup$
Oh ok now I get it. Didn't read properly I guess. Guess time to call it a day as well. Thanks again! ^-^
$endgroup$
– BCLC
Oct 25 '18 at 11:07











1












$begingroup$

$[0,1]$ is not connected, e.g. because $[0,frac12)$ is a non-trivial closed-and-open subset of it.



Not compact as ${1-frac1n: n=2,3,4,5,ldots}$ is an infinite subset without limit point in it.



Hausdorff because as you state its topology includes the usual, Hausdorff, one.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 12:08


















1












$begingroup$

$[0,1]$ is not connected, e.g. because $[0,frac12)$ is a non-trivial closed-and-open subset of it.



Not compact as ${1-frac1n: n=2,3,4,5,ldots}$ is an infinite subset without limit point in it.



Hausdorff because as you state its topology includes the usual, Hausdorff, one.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 12:08
















1












1








1





$begingroup$

$[0,1]$ is not connected, e.g. because $[0,frac12)$ is a non-trivial closed-and-open subset of it.



Not compact as ${1-frac1n: n=2,3,4,5,ldots}$ is an infinite subset without limit point in it.



Hausdorff because as you state its topology includes the usual, Hausdorff, one.






share|cite|improve this answer









$endgroup$



$[0,1]$ is not connected, e.g. because $[0,frac12)$ is a non-trivial closed-and-open subset of it.



Not compact as ${1-frac1n: n=2,3,4,5,ldots}$ is an infinite subset without limit point in it.



Hausdorff because as you state its topology includes the usual, Hausdorff, one.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 25 '18 at 11:55









Henno BrandsmaHenno Brandsma

108k347114




108k347114












  • $begingroup$
    $1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 12:08




















  • $begingroup$
    $1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
    $endgroup$
    – BCLC
    Oct 25 '18 at 12:08


















$begingroup$
$1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
$endgroup$
– BCLC
Oct 25 '18 at 12:08






$begingroup$
$1-frac1n$ was precisely my answer! I was afraid I was wrong somehow and that $1-frac1{2^n}$ is for some reason a right choice. Thanks ^-^
$endgroup$
– BCLC
Oct 25 '18 at 12:08




















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