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$(A_i)_iin E$ family of connected sets such that $bigcaplimits_iin E A_i neq emptyset$ then $bigcuplimits_i in E A_i$ is connected

Family of connected sets, proving union is connectedIf $A$ and $B$ are conneted and $Acap Bneq emptyset$, then $Acup B$ is connectedIf $E_alpha$ is connected, $bigcaplimits_alphain AE neq emptyset$, then $bigcuplimits_alphain AE$ is connectedHow to finish? ConnectednessIf $E$ and $F$ are connected subsets of $M$ with $Ecap Fneemptyset$, show that $Ecup F$ is connected.If $ Cl(A) cap B neq emptyset, E = A cup B$, prove that E is connected(X,$tau$) with $E_1,E_2 subseteq X$ connected and $E_1 cap Closure(E_2) neq emptyset$ implies $E_1 cup E_2$ connectedIntermediate value theorem (topology space)What about Union of connected sets?Prove a family of connected sets with one set intersecting all others is connected

0

$begingroup$

My idea so far is, since $bigcaplimits_iin E A_i neq emptyset$, then exists $p inbigcaplimits_iin E A_i$

if $bigcuplimits_i in E A_i$ is not connected, then exists $A$ and $B$ open sets such that $A bigcup B = bigcuplimits_i in E A_i$ with $A bigcap B = emptyset$. Then $p in A$ or $p in B$. I want to prove that if $p in A$ the $B = emptyset$

Thanks

connectedness

share|cite|improve this question

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

$endgroup$

add a comment | 

0

$begingroup$

My idea so far is, since $bigcaplimits_iin E A_i neq emptyset$, then exists $p inbigcaplimits_iin E A_i$

if $bigcuplimits_i in E A_i$ is not connected, then exists $A$ and $B$ open sets such that $A bigcup B = bigcuplimits_i in E A_i$ with $A bigcap B = emptyset$. Then $p in A$ or $p in B$. I want to prove that if $p in A$ the $B = emptyset$

Thanks

connectedness

share|cite|improve this question

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

$endgroup$

add a comment | 

$begingroup$

My idea so far is, since $bigcaplimits_iin E A_i neq emptyset$, then exists $p inbigcaplimits_iin E A_i$

if $bigcuplimits_i in E A_i$ is not connected, then exists $A$ and $B$ open sets such that $A bigcup B = bigcuplimits_i in E A_i$ with $A bigcap B = emptyset$. Then $p in A$ or $p in B$. I want to prove that if $p in A$ the $B = emptyset$

Thanks

connectedness

share|cite|improve this question

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

$endgroup$

share|cite|improve this question

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

share|cite|improve this question

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

edited Mar 10 at 19:49

rtybase

11.4k31533

edited Mar 10 at 19:49

rtybase

11.4k31533

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

asked Mar 10 at 19:17

Dani SeidlerDani Seidler

1839

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint:

Use that a topological space $(X,mathcalT)$ is connected iff every continuous function

$$f: (X, mathcalT) to (0,1, 2^0,1)$$ is constant.

share|cite|improve this answer

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

$endgroup$

add a comment | 

0

$begingroup$

Lemma: If $X$ is any topological space, $Zsubseteq X$ is both open and closed, $Vsubseteq X$ is a connected set then either $Vsubseteq Z$ or $Vsubseteq Xsetminus Z$.

I'll leave the proof of the lemma to you because it is almost trivial. Now let's use it for your exercise. Let $Zsubseteqcup_iin E A_i$ be a set which is both open and closed. For each $iin E$ the set $A_isubseteq cup_iin EA_i:=Y$ is connected and hence by the lemma $A_isubseteq Z$ or $A_isubseteq Ysetminus Z$. But now suppose that there are $i,jin E$ such that $A_isubseteq Z$ and $A_jsubseteq Ysetminus Z$. That means there is no element which is both in $A_i$ and $A_j$ which is a contradiction. So we conclude that either $A_isubseteq Z$ for all $iin E$ or $A_isubseteq Ysetminus Z$ for all $iin E$. If $A_isubseteq Z$ for all $iin E$ then the union is also a subset of $Z$ and hence $cup_iin EA_i=Z$. And if $A_isubseteq Ysetminus Z$ for all $iin E$ then $Z$ is empty.

share|cite|improve this answer

edited Mar 10 at 19:55

answered Mar 10 at 19:26

MarkMark

9,894622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it using the lemma. The way of proving the lemma is that if is false, then exists $v in partial Z$ but this either contradicts that $Z$ is open or that $V subseteq X setminus Z$
  $endgroup$
  – Dani Seidler
  Mar 10 at 21:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is even easier. Since $Z$ and $Xsetminus Z$ are both open sets in $X$ we conclude that $Vcap Z$ and $Vcap (Xsetminus Z)$ are both open in $V$. Since $V$ is connected one of them equals to $V$.
  $endgroup$
  – Mark
  Mar 10 at 22:07
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

Hint:

Use that a topological space $(X,mathcalT)$ is connected iff every continuous function

$$f: (X, mathcalT) to (0,1, 2^0,1)$$ is constant.

share|cite|improve this answer

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

$endgroup$

add a comment | 

0

$begingroup$

Hint:

Use that a topological space $(X,mathcalT)$ is connected iff every continuous function

$$f: (X, mathcalT) to (0,1, 2^0,1)$$ is constant.

share|cite|improve this answer

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

$endgroup$

add a comment | 

$begingroup$

Hint:

Use that a topological space $(X,mathcalT)$ is connected iff every continuous function

$$f: (X, mathcalT) to (0,1, 2^0,1)$$ is constant.

share|cite|improve this answer

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

$endgroup$

share|cite|improve this answer

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

answered Mar 10 at 19:54

Math_QEDMath_QED

7,70131453

0

$begingroup$

Lemma: If $X$ is any topological space, $Zsubseteq X$ is both open and closed, $Vsubseteq X$ is a connected set then either $Vsubseteq Z$ or $Vsubseteq Xsetminus Z$.

I'll leave the proof of the lemma to you because it is almost trivial. Now let's use it for your exercise. Let $Zsubseteqcup_iin E A_i$ be a set which is both open and closed. For each $iin E$ the set $A_isubseteq cup_iin EA_i:=Y$ is connected and hence by the lemma $A_isubseteq Z$ or $A_isubseteq Ysetminus Z$. But now suppose that there are $i,jin E$ such that $A_isubseteq Z$ and $A_jsubseteq Ysetminus Z$. That means there is no element which is both in $A_i$ and $A_j$ which is a contradiction. So we conclude that either $A_isubseteq Z$ for all $iin E$ or $A_isubseteq Ysetminus Z$ for all $iin E$. If $A_isubseteq Z$ for all $iin E$ then the union is also a subset of $Z$ and hence $cup_iin EA_i=Z$. And if $A_isubseteq Ysetminus Z$ for all $iin E$ then $Z$ is empty.

share|cite|improve this answer

edited Mar 10 at 19:55

answered Mar 10 at 19:26

MarkMark

9,894622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it using the lemma. The way of proving the lemma is that if is false, then exists $v in partial Z$ but this either contradicts that $Z$ is open or that $V subseteq X setminus Z$
  $endgroup$
  – Dani Seidler
  Mar 10 at 21:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is even easier. Since $Z$ and $Xsetminus Z$ are both open sets in $X$ we conclude that $Vcap Z$ and $Vcap (Xsetminus Z)$ are both open in $V$. Since $V$ is connected one of them equals to $V$.
  $endgroup$
  – Mark
  Mar 10 at 22:07
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Lemma: If $X$ is any topological space, $Zsubseteq X$ is both open and closed, $Vsubseteq X$ is a connected set then either $Vsubseteq Z$ or $Vsubseteq Xsetminus Z$.

I'll leave the proof of the lemma to you because it is almost trivial. Now let's use it for your exercise. Let $Zsubseteqcup_iin E A_i$ be a set which is both open and closed. For each $iin E$ the set $A_isubseteq cup_iin EA_i:=Y$ is connected and hence by the lemma $A_isubseteq Z$ or $A_isubseteq Ysetminus Z$. But now suppose that there are $i,jin E$ such that $A_isubseteq Z$ and $A_jsubseteq Ysetminus Z$. That means there is no element which is both in $A_i$ and $A_j$ which is a contradiction. So we conclude that either $A_isubseteq Z$ for all $iin E$ or $A_isubseteq Ysetminus Z$ for all $iin E$. If $A_isubseteq Z$ for all $iin E$ then the union is also a subset of $Z$ and hence $cup_iin EA_i=Z$. And if $A_isubseteq Ysetminus Z$ for all $iin E$ then $Z$ is empty.

share|cite|improve this answer

edited Mar 10 at 19:55

answered Mar 10 at 19:26

MarkMark

9,894622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I get it using the lemma. The way of proving the lemma is that if is false, then exists $v in partial Z$ but this either contradicts that $Z$ is open or that $V subseteq X setminus Z$
  $endgroup$
  – Dani Seidler
  Mar 10 at 21:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is even easier. Since $Z$ and $Xsetminus Z$ are both open sets in $X$ we conclude that $Vcap Z$ and $Vcap (Xsetminus Z)$ are both open in $V$. Since $V$ is connected one of them equals to $V$.
  $endgroup$
  – Mark
  Mar 10 at 22:07
  
  

  
  
  
  

add a comment | 

$begingroup$

Lemma: If $X$ is any topological space, $Zsubseteq X$ is both open and closed, $Vsubseteq X$ is a connected set then either $Vsubseteq Z$ or $Vsubseteq Xsetminus Z$.

I'll leave the proof of the lemma to you because it is almost trivial. Now let's use it for your exercise. Let $Zsubseteqcup_iin E A_i$ be a set which is both open and closed. For each $iin E$ the set $A_isubseteq cup_iin EA_i:=Y$ is connected and hence by the lemma $A_isubseteq Z$ or $A_isubseteq Ysetminus Z$. But now suppose that there are $i,jin E$ such that $A_isubseteq Z$ and $A_jsubseteq Ysetminus Z$. That means there is no element which is both in $A_i$ and $A_j$ which is a contradiction. So we conclude that either $A_isubseteq Z$ for all $iin E$ or $A_isubseteq Ysetminus Z$ for all $iin E$. If $A_isubseteq Z$ for all $iin E$ then the union is also a subset of $Z$ and hence $cup_iin EA_i=Z$. And if $A_isubseteq Ysetminus Z$ for all $iin E$ then $Z$ is empty.

share|cite|improve this answer

edited Mar 10 at 19:55

answered Mar 10 at 19:26

MarkMark

9,894622

$endgroup$

share|cite|improve this answer

edited Mar 10 at 19:55

answered Mar 10 at 19:26

MarkMark

9,894622

answered Mar 10 at 19:26

MarkMark

9,894622

answered Mar 10 at 19:26

MarkMark

9,894622