Given $X subset mathbbR^n$ connected, if $A subset X$ is such that $partial A cap X = emptyset$, then $A = emptyset$ or $A=X$

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Consider the statements:

(i) $X subset mathbbR^n$ is connected;

(ii) If $A subset X$ is such that $partial A cap X = emptyset$, then $A = emptyset$ or $A=X$.

Show that (i) implies (ii), but the converse is false.

Suppose $A subset X$ is such that $A neq emptyset$ and $A neq X$. Since $A subset X$, $A cup partial A = overlineA subset overlineX$. Thus, $(partial A cap overlineX) neq emptyset$, but $overlineX = X cup X'$, so $(partial A cap X) cup (partial A cap X') neq emptyset$.

That's all I got. My idea is to cover $X subset A_1 cup A_2$ so that $(partial A cap X)$ would be part of the set $A_1$ in order to use the connectedness. Any hint?

For the converse, I have no idea. I'm trying to get counterexamples in $mathbbR^2$ and $mathbbR^3$. I'm thinking first of a set that satisfies (ii), but in all cases I got a connected set.

real-analysis metric-spaces connectedness

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edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
  – GEdgar
  Jul 25 at 17:12
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

Consider the statements:

(i) $X subset mathbbR^n$ is connected;

(ii) If $A subset X$ is such that $partial A cap X = emptyset$, then $A = emptyset$ or $A=X$.

Show that (i) implies (ii), but the converse is false.

Suppose $A subset X$ is such that $A neq emptyset$ and $A neq X$. Since $A subset X$, $A cup partial A = overlineA subset overlineX$. Thus, $(partial A cap overlineX) neq emptyset$, but $overlineX = X cup X'$, so $(partial A cap X) cup (partial A cap X') neq emptyset$.

That's all I got. My idea is to cover $X subset A_1 cup A_2$ so that $(partial A cap X)$ would be part of the set $A_1$ in order to use the connectedness. Any hint?

For the converse, I have no idea. I'm trying to get counterexamples in $mathbbR^2$ and $mathbbR^3$. I'm thinking first of a set that satisfies (ii), but in all cases I got a connected set.

real-analysis metric-spaces connectedness

share|cite|improve this question

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
  – GEdgar
  Jul 25 at 17:12
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

Consider the statements:

(i) $X subset mathbbR^n$ is connected;

(ii) If $A subset X$ is such that $partial A cap X = emptyset$, then $A = emptyset$ or $A=X$.

Show that (i) implies (ii), but the converse is false.

Suppose $A subset X$ is such that $A neq emptyset$ and $A neq X$. Since $A subset X$, $A cup partial A = overlineA subset overlineX$. Thus, $(partial A cap overlineX) neq emptyset$, but $overlineX = X cup X'$, so $(partial A cap X) cup (partial A cap X') neq emptyset$.

That's all I got. My idea is to cover $X subset A_1 cup A_2$ so that $(partial A cap X)$ would be part of the set $A_1$ in order to use the connectedness. Any hint?

For the converse, I have no idea. I'm trying to get counterexamples in $mathbbR^2$ and $mathbbR^3$. I'm thinking first of a set that satisfies (ii), but in all cases I got a connected set.

real-analysis metric-spaces connectedness

share|cite|improve this question

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

Consider the statements:

(i) $X subset mathbbR^n$ is connected;

(ii) If $A subset X$ is such that $partial A cap X = emptyset$, then $A = emptyset$ or $A=X$.

Show that (i) implies (ii), but the converse is false.

Suppose $A subset X$ is such that $A neq emptyset$ and $A neq X$. Since $A subset X$, $A cup partial A = overlineA subset overlineX$. Thus, $(partial A cap overlineX) neq emptyset$, but $overlineX = X cup X'$, so $(partial A cap X) cup (partial A cap X') neq emptyset$.

That's all I got. My idea is to cover $X subset A_1 cup A_2$ so that $(partial A cap X)$ would be part of the set $A_1$ in order to use the connectedness. Any hint?

For the converse, I have no idea. I'm trying to get counterexamples in $mathbbR^2$ and $mathbbR^3$. I'm thinking first of a set that satisfies (ii), but in all cases I got a connected set.

real-analysis metric-spaces connectedness

share|cite|improve this question

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

edited Jul 25 at 17:27

Aloizio Macedo

22.5k23283

22.5k23283

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

asked Jul 25 at 17:07

Lucas Corrêa

1,106319

1,106319

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
  – GEdgar
  Jul 25 at 17:12
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
  – GEdgar
  Jul 25 at 17:12
  

  
  
  
  

1

1

Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
– GEdgar
Jul 25 at 17:12

Show $A$ is (relatively) closed in $X$; show the same for the complement $X setminus A$.
– GEdgar
Jul 25 at 17:12

add a comment | 

1 Answer
1

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oldest

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1
down vote



accepted

Assume that $X$ is connected and let $A subseteq X$ such that $partial A cap X = emptyset$. We will show that $A$ is both open and closed in $X$.

Recall that $overlineA = operatornameInt A cup partial A$.

We have $A = X cap operatornameInt A$. Indeed, clearly $X cap operatornameInt A subseteq X cap A = A$. Conversely, $$A = X cap A subseteq X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = X cap operatornameInt A$$
so we get that $A$ is open in $X$.

Also we have $A = X cap overlineA$ because

$$X cap overlineA = X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = A$$

so $A$ is closed in $X$.
If $A ne emptyset, X$ then $(A, Xsetminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

---

For a counterexample to the converse let $X$ be any disconnected closed subset of $mathbbR^n$. If $A subseteq X$ is such that $partial A cap X = emptyset$, we have $partial A subseteq overlineA subseteq overlineX = X$ so $partial A = partial A cap X = emptyset$.

It follows that $A$ is both open and closed in $mathbbR^n$. Indeed, we have $overlineA = operatornameInt A cup partial A = operatornameInt A$ so $operatornameInt A = A = overlineA$.
Since $mathbbR^n$ is connected, we conclude $A = emptyset$ or $A = mathbbR^n$, but the latter is impossible because $A subseteq X ne mathbbR^n$.

Hence $A = emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

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up vote
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accepted

Assume that $X$ is connected and let $A subseteq X$ such that $partial A cap X = emptyset$. We will show that $A$ is both open and closed in $X$.

Recall that $overlineA = operatornameInt A cup partial A$.

We have $A = X cap operatornameInt A$. Indeed, clearly $X cap operatornameInt A subseteq X cap A = A$. Conversely, $$A = X cap A subseteq X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = X cap operatornameInt A$$
so we get that $A$ is open in $X$.

Also we have $A = X cap overlineA$ because

$$X cap overlineA = X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = A$$

so $A$ is closed in $X$.
If $A ne emptyset, X$ then $(A, Xsetminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

---

For a counterexample to the converse let $X$ be any disconnected closed subset of $mathbbR^n$. If $A subseteq X$ is such that $partial A cap X = emptyset$, we have $partial A subseteq overlineA subseteq overlineX = X$ so $partial A = partial A cap X = emptyset$.

It follows that $A$ is both open and closed in $mathbbR^n$. Indeed, we have $overlineA = operatornameInt A cup partial A = operatornameInt A$ so $operatornameInt A = A = overlineA$.
Since $mathbbR^n$ is connected, we conclude $A = emptyset$ or $A = mathbbR^n$, but the latter is impossible because $A subseteq X ne mathbbR^n$.

Hence $A = emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote



accepted

Assume that $X$ is connected and let $A subseteq X$ such that $partial A cap X = emptyset$. We will show that $A$ is both open and closed in $X$.

Recall that $overlineA = operatornameInt A cup partial A$.

We have $A = X cap operatornameInt A$. Indeed, clearly $X cap operatornameInt A subseteq X cap A = A$. Conversely, $$A = X cap A subseteq X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = X cap operatornameInt A$$
so we get that $A$ is open in $X$.

Also we have $A = X cap overlineA$ because

$$X cap overlineA = X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = A$$

so $A$ is closed in $X$.
If $A ne emptyset, X$ then $(A, Xsetminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

---

For a counterexample to the converse let $X$ be any disconnected closed subset of $mathbbR^n$. If $A subseteq X$ is such that $partial A cap X = emptyset$, we have $partial A subseteq overlineA subseteq overlineX = X$ so $partial A = partial A cap X = emptyset$.

It follows that $A$ is both open and closed in $mathbbR^n$. Indeed, we have $overlineA = operatornameInt A cup partial A = operatornameInt A$ so $operatornameInt A = A = overlineA$.
Since $mathbbR^n$ is connected, we conclude $A = emptyset$ or $A = mathbbR^n$, but the latter is impossible because $A subseteq X ne mathbbR^n$.

Hence $A = emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Assume that $X$ is connected and let $A subseteq X$ such that $partial A cap X = emptyset$. We will show that $A$ is both open and closed in $X$.

Recall that $overlineA = operatornameInt A cup partial A$.

We have $A = X cap operatornameInt A$. Indeed, clearly $X cap operatornameInt A subseteq X cap A = A$. Conversely, $$A = X cap A subseteq X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = X cap operatornameInt A$$
so we get that $A$ is open in $X$.

Also we have $A = X cap overlineA$ because

$$X cap overlineA = X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = A$$

so $A$ is closed in $X$.
If $A ne emptyset, X$ then $(A, Xsetminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

---

For a counterexample to the converse let $X$ be any disconnected closed subset of $mathbbR^n$. If $A subseteq X$ is such that $partial A cap X = emptyset$, we have $partial A subseteq overlineA subseteq overlineX = X$ so $partial A = partial A cap X = emptyset$.

It follows that $A$ is both open and closed in $mathbbR^n$. Indeed, we have $overlineA = operatornameInt A cup partial A = operatornameInt A$ so $operatornameInt A = A = overlineA$.
Since $mathbbR^n$ is connected, we conclude $A = emptyset$ or $A = mathbbR^n$, but the latter is impossible because $A subseteq X ne mathbbR^n$.

Hence $A = emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

Assume that $X$ is connected and let $A subseteq X$ such that $partial A cap X = emptyset$. We will show that $A$ is both open and closed in $X$.

Recall that $overlineA = operatornameInt A cup partial A$.

We have $A = X cap operatornameInt A$. Indeed, clearly $X cap operatornameInt A subseteq X cap A = A$. Conversely, $$A = X cap A subseteq X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = X cap operatornameInt A$$
so we get that $A$ is open in $X$.

Also we have $A = X cap overlineA$ because

$$X cap overlineA = X cap (operatornameInt A cup partial A) = (X cap partial A) cup (X cap operatornameInt A) = A$$

so $A$ is closed in $X$.
If $A ne emptyset, X$ then $(A, Xsetminus A)$ would be a separation of $X$, which is impossible since $X$ is connected.

---

For a counterexample to the converse let $X$ be any disconnected closed subset of $mathbbR^n$. If $A subseteq X$ is such that $partial A cap X = emptyset$, we have $partial A subseteq overlineA subseteq overlineX = X$ so $partial A = partial A cap X = emptyset$.

It follows that $A$ is both open and closed in $mathbbR^n$. Indeed, we have $overlineA = operatornameInt A cup partial A = operatornameInt A$ so $operatornameInt A = A = overlineA$.
Since $mathbbR^n$ is connected, we conclude $A = emptyset$ or $A = mathbbR^n$, but the latter is impossible because $A subseteq X ne mathbbR^n$.

Hence $A = emptyset$ so $X$ satisfies the property $(2)$ but it isn't connected.

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 17:39

mechanodroid

22.2k52041

answered Jul 25 at 17:39

mechanodroid

22.2k52041

answered Jul 25 at 17:39

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment | 

 

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