Understanding the definition of a base of a metric space The 2019 Stack Overflow Developer Survey Results Are InDefinition of open and closed sets for metric spacesAny collection of subsets of $X$ can serve as a sub-base for a topologyDefinition of compact set/subsetShowing that for equivalent metrics, the separability of one metric space implies the separability of the otherIn a metric space with a countable base, how does every open cover has a countable subcover?Topology. Understanding what a base is intuitivelyA Base of a metric space intuitionExercise 2.23 in Baby RudinCountable base of a compact metric spaceDon't understand well the definition of compactness, or a compact set
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Understanding the definition of a base of a metric space
The 2019 Stack Overflow Developer Survey Results Are InDefinition of open and closed sets for metric spacesAny collection of subsets of $X$ can serve as a sub-base for a topologyDefinition of compact set/subsetShowing that for equivalent metrics, the separability of one metric space implies the separability of the otherIn a metric space with a countable base, how does every open cover has a countable subcover?Topology. Understanding what a base is intuitivelyA Base of a metric space intuitionExercise 2.23 in Baby RudinCountable base of a compact metric spaceDon't understand well the definition of compactness, or a compact set
$begingroup$
Definition: Let $(X,d)$ be a metric space.
A collection $v_n$ of subsets is said to be a base for $X$ if for every $x in X$ and every open set $G subset X$, such that $x in G$ we have $x in V_n subset G$ for some $N$.
The notation is what is giving me a hard time here. I just don't understand how it translates into "Every open set in $X$ is the union of a subcollection of $V_n$".
I don't understand how this is the case if that collection is always a subset of $G$, not equal to $G$?
real-analysis definition intuition
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
$endgroup$
add a comment |
$begingroup$
Definition: Let $(X,d)$ be a metric space.
A collection $v_n$ of subsets is said to be a base for $X$ if for every $x in X$ and every open set $G subset X$, such that $x in G$ we have $x in V_n subset G$ for some $N$.
The notation is what is giving me a hard time here. I just don't understand how it translates into "Every open set in $X$ is the union of a subcollection of $V_n$".
I don't understand how this is the case if that collection is always a subset of $G$, not equal to $G$?
real-analysis definition intuition
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
$endgroup$
add a comment |
$begingroup$
Definition: Let $(X,d)$ be a metric space.
A collection $v_n$ of subsets is said to be a base for $X$ if for every $x in X$ and every open set $G subset X$, such that $x in G$ we have $x in V_n subset G$ for some $N$.
The notation is what is giving me a hard time here. I just don't understand how it translates into "Every open set in $X$ is the union of a subcollection of $V_n$".
I don't understand how this is the case if that collection is always a subset of $G$, not equal to $G$?
real-analysis definition intuition
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
$endgroup$
Definition: Let $(X,d)$ be a metric space.
A collection $v_n$ of subsets is said to be a base for $X$ if for every $x in X$ and every open set $G subset X$, such that $x in G$ we have $x in V_n subset G$ for some $N$.
The notation is what is giving me a hard time here. I just don't understand how it translates into "Every open set in $X$ is the union of a subcollection of $V_n$".
I don't understand how this is the case if that collection is always a subset of $G$, not equal to $G$?
real-analysis definition intuition
real-analysis definition intuition
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
edited Mar 24 at 2:09
Viktor Glombik
1,3522628
1,3522628
asked Oct 23 '14 at 20:34
user186785user186785
212
asked Oct 23 '14 at 20:34
user186785user186785
212
asked Oct 23 '14 at 20:34
user186785user186785
212
212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take an open set $O$, and $xin O$. By definition, there exists $V=V_x$ in your base such that $xin Vsubseteq O$. This means that $O=bigcup x:xin Osubseteq bigcup V_xsubseteq O$; so $O=bigcup V_x$ is a union of open basic sets.
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
$endgroup$
add a comment |
$begingroup$
"Every open set in X is the union of a subcollection of $V_n$" means that any open set $U$ can be written as
$$U=bigcup V_i,$$
where $V_i$ belong to the base for any $i$. E.g. in $mathbbR$ with Euclidean topology, a base could be
$$B(x,r):x,rin mathbbQ.$$
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take an open set $O$, and $xin O$. By definition, there exists $V=V_x$ in your base such that $xin Vsubseteq O$. This means that $O=bigcup x:xin Osubseteq bigcup V_xsubseteq O$; so $O=bigcup V_x$ is a union of open basic sets.
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
$endgroup$
add a comment |
$begingroup$
Take an open set $O$, and $xin O$. By definition, there exists $V=V_x$ in your base such that $xin Vsubseteq O$. This means that $O=bigcup x:xin Osubseteq bigcup V_xsubseteq O$; so $O=bigcup V_x$ is a union of open basic sets.
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
$endgroup$
add a comment |
$begingroup$
Take an open set $O$, and $xin O$. By definition, there exists $V=V_x$ in your base such that $xin Vsubseteq O$. This means that $O=bigcup x:xin Osubseteq bigcup V_xsubseteq O$; so $O=bigcup V_x$ is a union of open basic sets.
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
$endgroup$
Take an open set $O$, and $xin O$. By definition, there exists $V=V_x$ in your base such that $xin Vsubseteq O$. This means that $O=bigcup x:xin Osubseteq bigcup V_xsubseteq O$; so $O=bigcup V_x$ is a union of open basic sets.
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
answered Oct 23 '14 at 20:37
Pedro Tamaroff♦Pedro Tamaroff
97.6k10153299
97.6k10153299
add a comment |
add a comment |
$begingroup$
"Every open set in X is the union of a subcollection of $V_n$" means that any open set $U$ can be written as
$$U=bigcup V_i,$$
where $V_i$ belong to the base for any $i$. E.g. in $mathbbR$ with Euclidean topology, a base could be
$$B(x,r):x,rin mathbbQ.$$
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
$endgroup$
add a comment |
$begingroup$
"Every open set in X is the union of a subcollection of $V_n$" means that any open set $U$ can be written as
$$U=bigcup V_i,$$
where $V_i$ belong to the base for any $i$. E.g. in $mathbbR$ with Euclidean topology, a base could be
$$B(x,r):x,rin mathbbQ.$$
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
$endgroup$
add a comment |
$begingroup$
"Every open set in X is the union of a subcollection of $V_n$" means that any open set $U$ can be written as
$$U=bigcup V_i,$$
where $V_i$ belong to the base for any $i$. E.g. in $mathbbR$ with Euclidean topology, a base could be
$$B(x,r):x,rin mathbbQ.$$
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
$endgroup$
"Every open set in X is the union of a subcollection of $V_n$" means that any open set $U$ can be written as
$$U=bigcup V_i,$$
where $V_i$ belong to the base for any $i$. E.g. in $mathbbR$ with Euclidean topology, a base could be
$$B(x,r):x,rin mathbbQ.$$
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
answered Oct 23 '14 at 20:37
MillyMilly
2,648612
2,648612
add a comment |
add a comment |
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