Connectedness and limit points in $mathbbR_l$

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I am studying general topology and currently going through the sections on connectedness of Munkres' book.

I am studying the connectedness of $mathbbR_l$. It is not connected as each interval $(-infty,a)$ and $[b,infty)$ is both open and closed. We can split each set by this logic until the only sets left are singletons, so those are the only connected sets.This means $mathbbR_l$ is totally disconnected. However, Lemma 23.1 says:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$

Then, if I take any separation of $mathbbR_l$ e.g. $(-infty,0) cup
[0,infty)$. I see then that the second set contains a limit point of the first one, so by Lemma 23.1 $mathbbR_l$ should be connected.

I would like to know why is not possible to use this Lemma on $mathbbR_l$

general-topology connectedness

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asked Aug 3 at 20:35

Amphiaraos

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I am studying general topology and currently going through the sections on connectedness of Munkres' book.

I am studying the connectedness of $mathbbR_l$. It is not connected as each interval $(-infty,a)$ and $[b,infty)$ is both open and closed. We can split each set by this logic until the only sets left are singletons, so those are the only connected sets.This means $mathbbR_l$ is totally disconnected. However, Lemma 23.1 says:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$

Then, if I take any separation of $mathbbR_l$ e.g. $(-infty,0) cup
[0,infty)$. I see then that the second set contains a limit point of the first one, so by Lemma 23.1 $mathbbR_l$ should be connected.

I would like to know why is not possible to use this Lemma on $mathbbR_l$

general-topology connectedness

share|cite|improve this question

asked Aug 3 at 20:35

Amphiaraos

12518

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am studying general topology and currently going through the sections on connectedness of Munkres' book.

I am studying the connectedness of $mathbbR_l$. It is not connected as each interval $(-infty,a)$ and $[b,infty)$ is both open and closed. We can split each set by this logic until the only sets left are singletons, so those are the only connected sets.This means $mathbbR_l$ is totally disconnected. However, Lemma 23.1 says:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$

Then, if I take any separation of $mathbbR_l$ e.g. $(-infty,0) cup
[0,infty)$. I see then that the second set contains a limit point of the first one, so by Lemma 23.1 $mathbbR_l$ should be connected.

I would like to know why is not possible to use this Lemma on $mathbbR_l$

general-topology connectedness

share|cite|improve this question

asked Aug 3 at 20:35

Amphiaraos

12518

I am studying general topology and currently going through the sections on connectedness of Munkres' book.

I am studying the connectedness of $mathbbR_l$. It is not connected as each interval $(-infty,a)$ and $[b,infty)$ is both open and closed. We can split each set by this logic until the only sets left are singletons, so those are the only connected sets.This means $mathbbR_l$ is totally disconnected. However, Lemma 23.1 says:

If $Y$ is a subspace of $X$, a separation of $Y$ is a pair of disjoint nonempty sets $A$ and $B$ whose union is $Y$, neither of which contains a limit point of the other. The space $Y$ is connected if there exists no separation of $Y$

Then, if I take any separation of $mathbbR_l$ e.g. $(-infty,0) cup
[0,infty)$. I see then that the second set contains a limit point of the first one, so by Lemma 23.1 $mathbbR_l$ should be connected.

I would like to know why is not possible to use this Lemma on $mathbbR_l$

general-topology connectedness

share|cite|improve this question

asked Aug 3 at 20:35

Amphiaraos

12518

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 20:35

Amphiaraos

12518

asked Aug 3 at 20:35

Amphiaraos

12518

asked Aug 3 at 20:35

Amphiaraos

12518

12518

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2 Answers
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In the topology on $mathbbR_l$, neither $(- infty, 0)$ or $[0, infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- infty, 0)$ because there is an open set containing $0$, namely $[0, infty)$, that doesn't intersect $(- infty, 0)$.

Sure, $0$ is a limit point of $(- infty, 0)$ in the Euclidean topology on $mathbbR$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.

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edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

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You are wrong: the second set contains no limit point of the first one. If $ain[0,+infty)$, then $[0,+infty)$ is an open set to which $a$ belongs and $[0,+infty)cap(-infty,0)=emptyset$. So, $a$ is not a limit point of $(-infty,0)$.

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answered Aug 3 at 20:42

José Carlos Santos

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

In the topology on $mathbbR_l$, neither $(- infty, 0)$ or $[0, infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- infty, 0)$ because there is an open set containing $0$, namely $[0, infty)$, that doesn't intersect $(- infty, 0)$.

Sure, $0$ is a limit point of $(- infty, 0)$ in the Euclidean topology on $mathbbR$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.

share|cite|improve this answer

edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

1,592210

add a comment | 

up vote
1
down vote



accepted

In the topology on $mathbbR_l$, neither $(- infty, 0)$ or $[0, infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- infty, 0)$ because there is an open set containing $0$, namely $[0, infty)$, that doesn't intersect $(- infty, 0)$.

Sure, $0$ is a limit point of $(- infty, 0)$ in the Euclidean topology on $mathbbR$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.

share|cite|improve this answer

edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

1,592210

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

In the topology on $mathbbR_l$, neither $(- infty, 0)$ or $[0, infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- infty, 0)$ because there is an open set containing $0$, namely $[0, infty)$, that doesn't intersect $(- infty, 0)$.

Sure, $0$ is a limit point of $(- infty, 0)$ in the Euclidean topology on $mathbbR$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.

share|cite|improve this answer

edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

1,592210

In the topology on $mathbbR_l$, neither $(- infty, 0)$ or $[0, infty)$ contains a limit point of the other. In particular, $0$ is not a limit point of $(- infty, 0)$ because there is an open set containing $0$, namely $[0, infty)$, that doesn't intersect $(- infty, 0)$.

Sure, $0$ is a limit point of $(- infty, 0)$ in the Euclidean topology on $mathbbR$. But why should that matter? The Euclidean topology isn't the topology we are dealing with.

share|cite|improve this answer

edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

1,592210

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 23:20

edited Aug 3 at 23:20

edited Aug 3 at 23:20

answered Aug 3 at 20:43

Chessanator

1,592210

answered Aug 3 at 20:43

Chessanator

1,592210

answered Aug 3 at 20:43

Chessanator

1,592210

1,592210

add a comment | 

add a comment | 

up vote
0
down vote

You are wrong: the second set contains no limit point of the first one. If $ain[0,+infty)$, then $[0,+infty)$ is an open set to which $a$ belongs and $[0,+infty)cap(-infty,0)=emptyset$. So, $a$ is not a limit point of $(-infty,0)$.

share|cite|improve this answer

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

add a comment | 

up vote
0
down vote

You are wrong: the second set contains no limit point of the first one. If $ain[0,+infty)$, then $[0,+infty)$ is an open set to which $a$ belongs and $[0,+infty)cap(-infty,0)=emptyset$. So, $a$ is not a limit point of $(-infty,0)$.

share|cite|improve this answer

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

add a comment | 

up vote
0
down vote

up vote
0
down vote

You are wrong: the second set contains no limit point of the first one. If $ain[0,+infty)$, then $[0,+infty)$ is an open set to which $a$ belongs and $[0,+infty)cap(-infty,0)=emptyset$. So, $a$ is not a limit point of $(-infty,0)$.

share|cite|improve this answer

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

You are wrong: the second set contains no limit point of the first one. If $ain[0,+infty)$, then $[0,+infty)$ is an open set to which $a$ belongs and $[0,+infty)cap(-infty,0)=emptyset$. So, $a$ is not a limit point of $(-infty,0)$.

share|cite|improve this answer

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

answered Aug 3 at 20:42

José Carlos Santos

112k1696172

112k1696172

add a comment | 

add a comment | 

 

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