Proof verification that standard topology is a topology
$begingroup$
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?
general-topology
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
add a comment |
$begingroup$
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?
general-topology
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
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
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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
add a comment |
$begingroup$
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?
general-topology
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
$endgroup$
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
general-topology
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
edited Jan 7 at 22:24
Matheus Andrade
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
asked Jan 7 at 22:11
Matheus AndradeMatheus Andrade
1,340418
1,340418
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
add a comment |
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
add a comment |
1 Answer
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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?
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
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add a comment |
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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?
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
$endgroup$
add a comment |
$begingroup$
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?
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
$endgroup$
add a comment |
$begingroup$
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?
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
$endgroup$
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?
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
edited Jan 8 at 3:29
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
answered Jan 8 at 3:21
William ElliotWilliam Elliot
8,9132820
8,9132820
add a comment |
add a comment |
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$begingroup$
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