Proof verification that standard topology is a topology












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Let $beta$ be the collection of all open intervals $(a,b) subset mathbb{R}$, with $a,b in mathbb{R}$. Prove that the topology $tau_beta$ generated by $beta$ is in fact a topology. Observation:



$tau_beta = {U subset mathbb{R} vert forall x in U ,exists B in beta text{ such that } x in B subset U}$




My proof: $emptyset in tau_beta$ vacuously. Since for every $x in mathbb{R}$ and $varepsilon > 0$, we have that $x in (x-varepsilon, x+varepsilon) subset mathbb{R}$, then $mathbb{R} in tau_beta$. Now, given any arbitrary $x in A = displaystyle{bigcup_{j in J} U_j} $, where $U_j$ is an open set for each $j$, then $x in U_i$ for some $i in J$. Since $U_i $ is open, then there exists an interval $B$ such that $x in B subset U_i$. Finally, consider $W = U_1 bigcap U_2$, where $U_1$ and $U_2$ are open sets. Given any $x in W$, since $x in U_1$ and $x in U_2$, then there exist $B_1, B_2 in beta$ such that $x in B_1 subset U_1$ and $x in B_2 subset U_2$. Then $x in B_1 bigcap B_2 subset U_1 bigcap U_2$ and we're done (the rest is trivial by induction).



Are all my steps correct?










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  • 1




    $begingroup$
    Yes it's correct. Maybe you should spell out the observation in the beginning..
    $endgroup$
    – Berci
    Jan 7 at 22:48












  • $begingroup$
    Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
    $endgroup$
    – Chris Custer
    Jan 8 at 3:42










  • $begingroup$
    This was an exercise given on my first day of a general topology summer course.
    $endgroup$
    – Matheus Andrade
    Jan 8 at 10:15










  • $begingroup$
    Ok. I guess it gets you to work through the definitions.
    $endgroup$
    – Chris Custer
    Jan 8 at 19:31
















1












$begingroup$



Let $beta$ be the collection of all open intervals $(a,b) subset mathbb{R}$, with $a,b in mathbb{R}$. Prove that the topology $tau_beta$ generated by $beta$ is in fact a topology. Observation:



$tau_beta = {U subset mathbb{R} vert forall x in U ,exists B in beta text{ such that } x in B subset U}$




My proof: $emptyset in tau_beta$ vacuously. Since for every $x in mathbb{R}$ and $varepsilon > 0$, we have that $x in (x-varepsilon, x+varepsilon) subset mathbb{R}$, then $mathbb{R} in tau_beta$. Now, given any arbitrary $x in A = displaystyle{bigcup_{j in J} U_j} $, where $U_j$ is an open set for each $j$, then $x in U_i$ for some $i in J$. Since $U_i $ is open, then there exists an interval $B$ such that $x in B subset U_i$. Finally, consider $W = U_1 bigcap U_2$, where $U_1$ and $U_2$ are open sets. Given any $x in W$, since $x in U_1$ and $x in U_2$, then there exist $B_1, B_2 in beta$ such that $x in B_1 subset U_1$ and $x in B_2 subset U_2$. Then $x in B_1 bigcap B_2 subset U_1 bigcap U_2$ and we're done (the rest is trivial by induction).



Are all my steps correct?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes it's correct. Maybe you should spell out the observation in the beginning..
    $endgroup$
    – Berci
    Jan 7 at 22:48












  • $begingroup$
    Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
    $endgroup$
    – Chris Custer
    Jan 8 at 3:42










  • $begingroup$
    This was an exercise given on my first day of a general topology summer course.
    $endgroup$
    – Matheus Andrade
    Jan 8 at 10:15










  • $begingroup$
    Ok. I guess it gets you to work through the definitions.
    $endgroup$
    – Chris Custer
    Jan 8 at 19:31














1












1








1





$begingroup$



Let $beta$ be the collection of all open intervals $(a,b) subset mathbb{R}$, with $a,b in mathbb{R}$. Prove that the topology $tau_beta$ generated by $beta$ is in fact a topology. Observation:



$tau_beta = {U subset mathbb{R} vert forall x in U ,exists B in beta text{ such that } x in B subset U}$




My proof: $emptyset in tau_beta$ vacuously. Since for every $x in mathbb{R}$ and $varepsilon > 0$, we have that $x in (x-varepsilon, x+varepsilon) subset mathbb{R}$, then $mathbb{R} in tau_beta$. Now, given any arbitrary $x in A = displaystyle{bigcup_{j in J} U_j} $, where $U_j$ is an open set for each $j$, then $x in U_i$ for some $i in J$. Since $U_i $ is open, then there exists an interval $B$ such that $x in B subset U_i$. Finally, consider $W = U_1 bigcap U_2$, where $U_1$ and $U_2$ are open sets. Given any $x in W$, since $x in U_1$ and $x in U_2$, then there exist $B_1, B_2 in beta$ such that $x in B_1 subset U_1$ and $x in B_2 subset U_2$. Then $x in B_1 bigcap B_2 subset U_1 bigcap U_2$ and we're done (the rest is trivial by induction).



Are all my steps correct?










share|cite|improve this question











$endgroup$





Let $beta$ be the collection of all open intervals $(a,b) subset mathbb{R}$, with $a,b in mathbb{R}$. Prove that the topology $tau_beta$ generated by $beta$ is in fact a topology. Observation:



$tau_beta = {U subset mathbb{R} vert forall x in U ,exists B in beta text{ such that } x in B subset U}$




My proof: $emptyset in tau_beta$ vacuously. Since for every $x in mathbb{R}$ and $varepsilon > 0$, we have that $x in (x-varepsilon, x+varepsilon) subset mathbb{R}$, then $mathbb{R} in tau_beta$. Now, given any arbitrary $x in A = displaystyle{bigcup_{j in J} U_j} $, where $U_j$ is an open set for each $j$, then $x in U_i$ for some $i in J$. Since $U_i $ is open, then there exists an interval $B$ such that $x in B subset U_i$. Finally, consider $W = U_1 bigcap U_2$, where $U_1$ and $U_2$ are open sets. Given any $x in W$, since $x in U_1$ and $x in U_2$, then there exist $B_1, B_2 in beta$ such that $x in B_1 subset U_1$ and $x in B_2 subset U_2$. Then $x in B_1 bigcap B_2 subset U_1 bigcap U_2$ and we're done (the rest is trivial by induction).



Are all my steps correct?







general-topology






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edited Jan 7 at 22:24







Matheus Andrade

















asked Jan 7 at 22:11









Matheus AndradeMatheus Andrade

1,390418




1,390418








  • 1




    $begingroup$
    Yes it's correct. Maybe you should spell out the observation in the beginning..
    $endgroup$
    – Berci
    Jan 7 at 22:48












  • $begingroup$
    Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
    $endgroup$
    – Chris Custer
    Jan 8 at 3:42










  • $begingroup$
    This was an exercise given on my first day of a general topology summer course.
    $endgroup$
    – Matheus Andrade
    Jan 8 at 10:15










  • $begingroup$
    Ok. I guess it gets you to work through the definitions.
    $endgroup$
    – Chris Custer
    Jan 8 at 19:31














  • 1




    $begingroup$
    Yes it's correct. Maybe you should spell out the observation in the beginning..
    $endgroup$
    – Berci
    Jan 7 at 22:48












  • $begingroup$
    Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
    $endgroup$
    – Chris Custer
    Jan 8 at 3:42










  • $begingroup$
    This was an exercise given on my first day of a general topology summer course.
    $endgroup$
    – Matheus Andrade
    Jan 8 at 10:15










  • $begingroup$
    Ok. I guess it gets you to work through the definitions.
    $endgroup$
    – Chris Custer
    Jan 8 at 19:31








1




1




$begingroup$
Yes it's correct. Maybe you should spell out the observation in the beginning..
$endgroup$
– Berci
Jan 7 at 22:48






$begingroup$
Yes it's correct. Maybe you should spell out the observation in the beginning..
$endgroup$
– Berci
Jan 7 at 22:48














$begingroup$
Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
$endgroup$
– Chris Custer
Jan 8 at 3:42




$begingroup$
Isn't the topology generated by a subset of $P(A)$ always a topology on $A$? (See @William Elliot's answer). Out of curiosity, where is this problem from?
$endgroup$
– Chris Custer
Jan 8 at 3:42












$begingroup$
This was an exercise given on my first day of a general topology summer course.
$endgroup$
– Matheus Andrade
Jan 8 at 10:15




$begingroup$
This was an exercise given on my first day of a general topology summer course.
$endgroup$
– Matheus Andrade
Jan 8 at 10:15












$begingroup$
Ok. I guess it gets you to work through the definitions.
$endgroup$
– Chris Custer
Jan 8 at 19:31




$begingroup$
Ok. I guess it gets you to work through the definitions.
$endgroup$
– Chris Custer
Jan 8 at 19:31










1 Answer
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That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.



Exercise. If T is a not empty collection of topologies for X,

show $cap$T is a topology for X.



Problem. Is there a topology for X that contains any subset of P(X)?

Why is this important?






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    $begingroup$

    That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.



    Exercise. If T is a not empty collection of topologies for X,

    show $cap$T is a topology for X.



    Problem. Is there a topology for X that contains any subset of P(X)?

    Why is this important?






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.



      Exercise. If T is a not empty collection of topologies for X,

      show $cap$T is a topology for X.



      Problem. Is there a topology for X that contains any subset of P(X)?

      Why is this important?






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.



        Exercise. If T is a not empty collection of topologies for X,

        show $cap$T is a topology for X.



        Problem. Is there a topology for X that contains any subset of P(X)?

        Why is this important?






        share|cite|improve this answer











        $endgroup$



        That is a vacuous problem because the topology generated by any subset K of P(X) is by definition a topology for X, the smallest topology for X containing K, the intersection of all topologies for X containing K.



        Exercise. If T is a not empty collection of topologies for X,

        show $cap$T is a topology for X.



        Problem. Is there a topology for X that contains any subset of P(X)?

        Why is this important?







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 3:29

























        answered Jan 8 at 3:21









        William ElliotWilliam Elliot

        9,1812820




        9,1812820






























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