Prove every subset of in the discrete metric is clopen Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)discrete metric, both open and closed.discrete metric, both open and closed.Open set in a metric space is union of closed balls?Every metric space contains a discrete, coarsely dense subsetProving that a subset endowed with the discrete metric is both open and closed - choice of radius of the ball around a pointIn a metric space, is every open set the countable union of closed sets?Compact Sets of $(X,d)$ with discrete metricDoes there exist any non discrete metric space $(X,d)$ in which every $F_sigma$ (resp. $G_delta$) set is clopen?Open and closed balls in discrete metricOpen and Closed Sets Discrete MetricHow to prove the set of bounded sequences is clopen in the uniform metric?
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Prove every subset of in the discrete metric is clopen
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)discrete metric, both open and closed.discrete metric, both open and closed.Open set in a metric space is union of closed balls?Every metric space contains a discrete, coarsely dense subsetProving that a subset endowed with the discrete metric is both open and closed - choice of radius of the ball around a pointIn a metric space, is every open set the countable union of closed sets?Compact Sets of $(X,d)$ with discrete metricDoes there exist any non discrete metric space $(X,d)$ in which every $F_sigma$ (resp. $G_delta$) set is clopen?Open and closed balls in discrete metricOpen and Closed Sets Discrete MetricHow to prove the set of bounded sequences is clopen in the uniform metric?
$begingroup$
Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:
”Given a set $X$ and metric $d(x, y) = 1$ if $x neq y$ and $d(x, y) = 0$ if $x = y$ then we want to prove that every subset of the resulting metric space $(X, d)$ is both open and closed.”.
And the solution is as follows:
”Since each ball $B(x; frac12)$ reduces to the singleton set $x$, every subset is a union of open balls, hence every subset is open.”.
My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.
In the book there is a theorem that states that a subset of $X$ is open if and only if it is a union of open balls in $X$, and is being used in the proof.
I get that in $X$ each subset is a singleton set $x$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $B(x; frac12)$ then each collection of singletons can be written as a union of these open balls and thus each subset of $X$ is open.
But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.
How do I formalize? Any feedback is greatly appreciated. Thanks in advance.
/Isak
metric-spaces
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
iaenstromiaenstrom
487
$endgroup$
add a comment |
$begingroup$
Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:
”Given a set $X$ and metric $d(x, y) = 1$ if $x neq y$ and $d(x, y) = 0$ if $x = y$ then we want to prove that every subset of the resulting metric space $(X, d)$ is both open and closed.”.
And the solution is as follows:
”Since each ball $B(x; frac12)$ reduces to the singleton set $x$, every subset is a union of open balls, hence every subset is open.”.
My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.
In the book there is a theorem that states that a subset of $X$ is open if and only if it is a union of open balls in $X$, and is being used in the proof.
I get that in $X$ each subset is a singleton set $x$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $B(x; frac12)$ then each collection of singletons can be written as a union of these open balls and thus each subset of $X$ is open.
But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.
How do I formalize? Any feedback is greatly appreciated. Thanks in advance.
/Isak
metric-spaces
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
iaenstromiaenstrom
487
$endgroup$
1
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
1
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday
add a comment |
$begingroup$
Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:
”Given a set $X$ and metric $d(x, y) = 1$ if $x neq y$ and $d(x, y) = 0$ if $x = y$ then we want to prove that every subset of the resulting metric space $(X, d)$ is both open and closed.”.
And the solution is as follows:
”Since each ball $B(x; frac12)$ reduces to the singleton set $x$, every subset is a union of open balls, hence every subset is open.”.
My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.
In the book there is a theorem that states that a subset of $X$ is open if and only if it is a union of open balls in $X$, and is being used in the proof.
I get that in $X$ each subset is a singleton set $x$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $B(x; frac12)$ then each collection of singletons can be written as a union of these open balls and thus each subset of $X$ is open.
But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.
How do I formalize? Any feedback is greatly appreciated. Thanks in advance.
/Isak
metric-spaces
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
iaenstromiaenstrom
487
$endgroup$
Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:
”Given a set $X$ and metric $d(x, y) = 1$ if $x neq y$ and $d(x, y) = 0$ if $x = y$ then we want to prove that every subset of the resulting metric space $(X, d)$ is both open and closed.”.
And the solution is as follows:
”Since each ball $B(x; frac12)$ reduces to the singleton set $x$, every subset is a union of open balls, hence every subset is open.”.
My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.
In the book there is a theorem that states that a subset of $X$ is open if and only if it is a union of open balls in $X$, and is being used in the proof.
I get that in $X$ each subset is a singleton set $x$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $B(x; frac12)$ then each collection of singletons can be written as a union of these open balls and thus each subset of $X$ is open.
But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.
How do I formalize? Any feedback is greatly appreciated. Thanks in advance.
/Isak
metric-spaces
metric-spaces
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
iaenstromiaenstrom
487
edited yesterday
YuiTo Cheng
2,43841037
asked yesterday
iaenstromiaenstrom
487
edited yesterday
YuiTo Cheng
2,43841037
edited yesterday
YuiTo Cheng
2,43841037
edited yesterday
YuiTo Cheng
2,43841037
2,43841037
asked yesterday
iaenstromiaenstrom
487
asked yesterday
iaenstromiaenstrom
487
asked yesterday
iaenstromiaenstrom
487
487
1
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
1
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday
add a comment |
1
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
1
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday
1
1
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
1
1
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let every subset of a topolgical space be open. A subset is closed if and only if its complement is open. The complement of every subset is a subset so is open. Therefore every subset is closed
answered yesterday
G AkerG Aker
3237
$endgroup$
add a comment |
$begingroup$
Another way to reason: suppose $x in overlineA$ (the closure of $A$) for some arbitary subset of $A$. Then every ball around $x$ intersects $A$, in particular $B(x,frac12)=x$ must intersect $A$, which means $x in A$.
So for all $A subseteq X$, $overlineA subseteq A (subseteq overlineA)$ so $A = overlineA$ and every subset $A$ is closed.
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let every subset of a topolgical space be open. A subset is closed if and only if its complement is open. The complement of every subset is a subset so is open. Therefore every subset is closed
answered yesterday
G AkerG Aker
3237
$endgroup$
add a comment |
$begingroup$
Let every subset of a topolgical space be open. A subset is closed if and only if its complement is open. The complement of every subset is a subset so is open. Therefore every subset is closed
answered yesterday
G AkerG Aker
3237
$endgroup$
add a comment |
$begingroup$
Let every subset of a topolgical space be open. A subset is closed if and only if its complement is open. The complement of every subset is a subset so is open. Therefore every subset is closed
answered yesterday
G AkerG Aker
3237
$endgroup$
Let every subset of a topolgical space be open. A subset is closed if and only if its complement is open. The complement of every subset is a subset so is open. Therefore every subset is closed
answered yesterday
G AkerG Aker
3237
answered yesterday
G AkerG Aker
3237
answered yesterday
G AkerG Aker
3237
answered yesterday
G AkerG Aker
3237
3237
add a comment |
add a comment |
$begingroup$
Another way to reason: suppose $x in overlineA$ (the closure of $A$) for some arbitary subset of $A$. Then every ball around $x$ intersects $A$, in particular $B(x,frac12)=x$ must intersect $A$, which means $x in A$.
So for all $A subseteq X$, $overlineA subseteq A (subseteq overlineA)$ so $A = overlineA$ and every subset $A$ is closed.
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
Another way to reason: suppose $x in overlineA$ (the closure of $A$) for some arbitary subset of $A$. Then every ball around $x$ intersects $A$, in particular $B(x,frac12)=x$ must intersect $A$, which means $x in A$.
So for all $A subseteq X$, $overlineA subseteq A (subseteq overlineA)$ so $A = overlineA$ and every subset $A$ is closed.
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
Another way to reason: suppose $x in overlineA$ (the closure of $A$) for some arbitary subset of $A$. Then every ball around $x$ intersects $A$, in particular $B(x,frac12)=x$ must intersect $A$, which means $x in A$.
So for all $A subseteq X$, $overlineA subseteq A (subseteq overlineA)$ so $A = overlineA$ and every subset $A$ is closed.
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
Another way to reason: suppose $x in overlineA$ (the closure of $A$) for some arbitary subset of $A$. Then every ball around $x$ intersects $A$, in particular $B(x,frac12)=x$ must intersect $A$, which means $x in A$.
So for all $A subseteq X$, $overlineA subseteq A (subseteq overlineA)$ so $A = overlineA$ and every subset $A$ is closed.
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
answered yesterday
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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1
$begingroup$
Possible duplicate of discrete metric, both open and closed.
$endgroup$
– YuiTo Cheng
yesterday
1
$begingroup$
You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves.
$endgroup$
– Ben Millwood
yesterday