Definition of Adherent Point
I am using the following definition:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if its neighborhood Br(x) contains an element in A for all r>0.
For one of my exercise, it would be really useful if I can use another definition of adherent point:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if every open set in X containing x contains an element in A.
How can I show these two definitions are equivalent?
real-analysis general-topology
asked Nov 26 at 3:31
Fluffy Skye
778
add a comment |
I am using the following definition:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if its neighborhood Br(x) contains an element in A for all r>0.
For one of my exercise, it would be really useful if I can use another definition of adherent point:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if every open set in X containing x contains an element in A.
How can I show these two definitions are equivalent?
real-analysis general-topology
asked Nov 26 at 3:31
Fluffy Skye
778
Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22
add a comment |
I am using the following definition:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if its neighborhood Br(x) contains an element in A for all r>0.
For one of my exercise, it would be really useful if I can use another definition of adherent point:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if every open set in X containing x contains an element in A.
How can I show these two definitions are equivalent?
real-analysis general-topology
asked Nov 26 at 3:31
Fluffy Skye
778
I am using the following definition:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if its neighborhood Br(x) contains an element in A for all r>0.
For one of my exercise, it would be really useful if I can use another definition of adherent point:
(X,d) a metric space, A is a subset of X,
for any x in X, it is an adherent point of A if every open set in X containing x contains an element in A.
How can I show these two definitions are equivalent?
real-analysis general-topology
real-analysis general-topology
asked Nov 26 at 3:31
Fluffy Skye
778
asked Nov 26 at 3:31
Fluffy Skye
778
asked Nov 26 at 3:31
Fluffy Skye
778
asked Nov 26 at 3:31
Fluffy Skye
778
asked Nov 26 at 3:31
Fluffy Skye
778
778
Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22
add a comment |
Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22
Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22
Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22
add a comment |
1 Answer
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Those are equivalent in metric spaces: for every open $O$ that contains $x$, there
is some $r>0$ such that $B(x,r) subseteq O$ ($O$ is a union of balls, or the definition of open sets in a metric space). So if every open ball around $x$ intersects $A$, so does every open set containing $x$.
The reverse also holds, as open balls are themselves in particular open sets in the topology.
answered Nov 26 at 6:15
Henno Brandsma
104k346113
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
Those are equivalent in metric spaces: for every open $O$ that contains $x$, there
is some $r>0$ such that $B(x,r) subseteq O$ ($O$ is a union of balls, or the definition of open sets in a metric space). So if every open ball around $x$ intersects $A$, so does every open set containing $x$.
The reverse also holds, as open balls are themselves in particular open sets in the topology.
answered Nov 26 at 6:15
Henno Brandsma
104k346113
add a comment |
Those are equivalent in metric spaces: for every open $O$ that contains $x$, there
is some $r>0$ such that $B(x,r) subseteq O$ ($O$ is a union of balls, or the definition of open sets in a metric space). So if every open ball around $x$ intersects $A$, so does every open set containing $x$.
The reverse also holds, as open balls are themselves in particular open sets in the topology.
answered Nov 26 at 6:15
Henno Brandsma
104k346113
add a comment |
Those are equivalent in metric spaces: for every open $O$ that contains $x$, there
is some $r>0$ such that $B(x,r) subseteq O$ ($O$ is a union of balls, or the definition of open sets in a metric space). So if every open ball around $x$ intersects $A$, so does every open set containing $x$.
The reverse also holds, as open balls are themselves in particular open sets in the topology.
answered Nov 26 at 6:15
Henno Brandsma
104k346113
Those are equivalent in metric spaces: for every open $O$ that contains $x$, there
is some $r>0$ such that $B(x,r) subseteq O$ ($O$ is a union of balls, or the definition of open sets in a metric space). So if every open ball around $x$ intersects $A$, so does every open set containing $x$.
The reverse also holds, as open balls are themselves in particular open sets in the topology.
answered Nov 26 at 6:15
Henno Brandsma
104k346113
answered Nov 26 at 6:15
Henno Brandsma
104k346113
answered Nov 26 at 6:15
Henno Brandsma
104k346113
answered Nov 26 at 6:15
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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Use the fact that every open set is a union of balls.
– William Elliot
Nov 26 at 4:22