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Visualization / sketch for this basic proof about subspace topology

In a metric space, closure is closedChain of compact subsets and their intersectionOpen set with respect to french railway metricFor compact $K$ and open $U supseteq K$, there exists $varepsilon>0$ such that $B(K,varepsilon) subseteq U$Open or closed set in $mathbb{R}$Proof regarding nested compact subsets of a metric spaceProve that closed and bounded subsets of metric spaces are compactThe fact that subspace topology coincides with induced metric topology leads to a contradictionA curiosity on covers of compactIntersection of neighborhoods of closed subsets in a metric space

0

$begingroup$

Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.

My proof of "$implies$" is pretty straight forward:

Let $Bsubset A$ be an open subset.
Define
begin{equation*}
U_{varepsilon}^{Z}(x)
:= { y in Z: d_A(x,y) = d(x,y) < varepsilon}
end{equation*}
for any subset $Z in X$.
Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
begin{equation*}
Bsupset U^{A}_{varepsilon_{x}}(x)
= Acap U^{X}_{varepsilon_{x}}(x).
end{equation*}
The set
begin{equation*}
Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
end{equation*}
is open in $X$ because it's the union of open sets in $X$.
Furthermore we have $B = Acap Y$.

But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.

Any help is greatly appreciated :)

general-topology metric-spaces visualization

share|cite|improve this question

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

$endgroup$

add a comment | 

0

$begingroup$

Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.

My proof of "$implies$" is pretty straight forward:

Let $Bsubset A$ be an open subset.
Define
begin{equation*}
U_{varepsilon}^{Z}(x)
:= { y in Z: d_A(x,y) = d(x,y) < varepsilon}
end{equation*}
for any subset $Z in X$.
Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
begin{equation*}
Bsupset U^{A}_{varepsilon_{x}}(x)
= Acap U^{X}_{varepsilon_{x}}(x).
end{equation*}
The set
begin{equation*}
Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
end{equation*}
is open in $X$ because it's the union of open sets in $X$.
Furthermore we have $B = Acap Y$.

But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.

Any help is greatly appreciated :)

general-topology metric-spaces visualization

share|cite|improve this question

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

$endgroup$

add a comment | 

$begingroup$

Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.

My proof of "$implies$" is pretty straight forward:

Let $Bsubset A$ be an open subset.
Define
begin{equation*}
U_{varepsilon}^{Z}(x)
:= { y in Z: d_A(x,y) = d(x,y) < varepsilon}
end{equation*}
for any subset $Z in X$.
Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
begin{equation*}
Bsupset U^{A}_{varepsilon_{x}}(x)
= Acap U^{X}_{varepsilon_{x}}(x).
end{equation*}
The set
begin{equation*}
Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
end{equation*}
is open in $X$ because it's the union of open sets in $X$.
Furthermore we have $B = Acap Y$.

But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.

Any help is greatly appreciated :)

general-topology metric-spaces visualization

share|cite|improve this question

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

$endgroup$

share|cite|improve this question

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

share|cite|improve this question

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

asked 2 days ago

Viktor GlombikViktor Glombik

1,0121528

1 Answer
1

active

oldest

votes

1

$begingroup$

Draw $X$ to be the plane and $A$ to be the $x$-axis.

share|cite|improve this answer

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, but when I want to draw a sketch, an interval is more convenient, right?
  $endgroup$
  – Viktor Glombik
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

Draw $X$ to be the plane and $A$ to be the $x$-axis.

share|cite|improve this answer

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, but when I want to draw a sketch, an interval is more convenient, right?
  $endgroup$
  – Viktor Glombik
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Draw $X$ to be the plane and $A$ to be the $x$-axis.

share|cite|improve this answer

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
  $endgroup$
  – Viktor Glombik
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, but when I want to draw a sketch, an interval is more convenient, right?
  $endgroup$
  – Viktor Glombik
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
  $endgroup$
  – Lee Mosher
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Draw $X$ to be the plane and $A$ to be the $x$-axis.

share|cite|improve this answer

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

$endgroup$

share|cite|improve this answer

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

answered 2 days ago

Lee MosherLee Mosher

50.3k33787

answered 2 days ago

Lee MosherLee Mosher

50.3k33787