Baby Rudin Definition 2.18

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This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:

Now, Consider the interval (1,2).

This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.

Open at the same time according to the $(e), (f)$ given in the definition.

Please clarify.

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

4

Please do not delete questions after having gotten an answer.
– quid♦
Nov 17 at 14:38

There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
Nov 17 at 15:36

... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
Nov 17 at 16:11

@AccidentalFourierTransform can you please give one example except $mathbbR$?
– HindShah
Nov 17 at 17:16

In the metric space $[0,1] cup 2$ with the inherited metric from the reals, $2$ is open and closed, e.g. $N_1(2) = 2$ so it's open, and $2$ has no limit points, so it's closed.
– Henno Brandsma
2 days ago

add a comment |

up vote
0
down vote

favorite

This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:

Now, Consider the interval (1,2).

This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.

Open at the same time according to the $(e), (f)$ given in the definition.

Please clarify.

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

4

Please do not delete questions after having gotten an answer.
– quid♦
Nov 17 at 14:38

There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
Nov 17 at 15:36

... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
Nov 17 at 16:11

@AccidentalFourierTransform can you please give one example except $mathbbR$?
– HindShah
Nov 17 at 17:16

In the metric space $[0,1] cup 2$ with the inherited metric from the reals, $2$ is open and closed, e.g. $N_1(2) = 2$ so it's open, and $2$ has no limit points, so it's closed.
– Henno Brandsma
2 days ago

add a comment |

up vote
0
down vote

favorite

This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:

Now, Consider the interval (1,2).

This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.

Open at the same time according to the $(e), (f)$ given in the definition.

Please clarify.

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

This are few Definitions stated in Principles of Mathematical Analysis by Walter Rudin:

Now, Consider the interval (1,2).

This interval is closed because according to $(b)$ the every point of this set contains another point in its neighbourhood, thus a limit point, hence closed.

Open at the same time according to the $(e), (f)$ given in the definition.

Please clarify.

general-topology metric-spaces

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

edited Nov 17 at 15:35

Henno Brandsma

101k344107

edited Nov 17 at 15:35

Henno Brandsma

101k344107

edited Nov 17 at 15:35

Henno Brandsma

101k344107

asked Nov 17 at 12:42

HindShah

New contributor

asked Nov 17 at 12:42

HindShah

asked Nov 17 at 12:42

HindShah

New contributor

HindShah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

4

Please do not delete questions after having gotten an answer.
– quid♦
Nov 17 at 14:38

There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
Nov 17 at 15:36

... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
Nov 17 at 16:11

@AccidentalFourierTransform can you please give one example except $mathbbR$?
– HindShah
Nov 17 at 17:16

In the metric space $[0,1] cup 2$ with the inherited metric from the reals, $2$ is open and closed, e.g. $N_1(2) = 2$ so it's open, and $2$ has no limit points, so it's closed.
– Henno Brandsma
2 days ago

add a comment |

4

Please do not delete questions after having gotten an answer.
– quid♦
Nov 17 at 14:38

There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
Nov 17 at 15:36

... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
Nov 17 at 16:11

@AccidentalFourierTransform can you please give one example except $mathbbR$?
– HindShah
Nov 17 at 17:16

In the metric space $[0,1] cup 2$ with the inherited metric from the reals, $2$ is open and closed, e.g. $N_1(2) = 2$ so it's open, and $2$ has no limit points, so it's closed.
– Henno Brandsma
2 days ago

Please do not delete questions after having gotten an answer.
– quid♦
Nov 17 at 14:38

There is no problem in itself with a set being open and closed. That can actually happen too.
– Henno Brandsma
Nov 17 at 15:36

... also, sets can be open and closed at the same time.
– AccidentalFourierTransform
Nov 17 at 16:11

@AccidentalFourierTransform can you please give one example except $mathbbR$?
– HindShah
Nov 17 at 17:16

In the metric space $[0,1] cup 2$ with the inherited metric from the reals, $2$ is open and closed, e.g. $N_1(2) = 2$ so it's open, and $2$ has no limit points, so it's closed.
– Henno Brandsma
2 days ago

add a comment |

2 Answers
2

active

oldest

votes

up vote
3
down vote

accepted

There is a difference between

Every limit point of $E$ is a point of $E$ . . . . .$(1)$

and

Every point of $E$ is a limit point of $E$ . . . . . .$(2)$

You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.

And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

add a comment |

up vote
3
down vote

$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.

answered Nov 17 at 12:48

Henno Brandsma

101k344107

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

accepted

There is a difference between

Every limit point of $E$ is a point of $E$ . . . . .$(1)$

and

Every point of $E$ is a limit point of $E$ . . . . . .$(2)$

You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.

And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

add a comment |

up vote
3
down vote

accepted

There is a difference between

Every limit point of $E$ is a point of $E$ . . . . .$(1)$

and

Every point of $E$ is a limit point of $E$ . . . . . .$(2)$

You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.

And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

add a comment |

up vote
3
down vote

accepted

There is a difference between

Every limit point of $E$ is a point of $E$ . . . . .$(1)$

and

Every point of $E$ is a limit point of $E$ . . . . . .$(2)$

You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.

And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

There is a difference between

Every limit point of $E$ is a point of $E$ . . . . .$(1)$

and

Every point of $E$ is a limit point of $E$ . . . . . .$(2)$

You have reasoned that $(2)$ is true and concluded that $E$ is closed. But this is incorrect, because you need to check whether $(1)$ is true, since $E$ is defined to be closed if $(1)$ holds.

And indeed, $1$ is a limit point of $E = (1,2)$ but $1 notin E$, so $E$ is not closed.

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

edited Nov 17 at 16:33

answered Nov 17 at 15:19

Brahadeesh

5,53941956

answered Nov 17 at 15:19

Brahadeesh

5,53941956

answered Nov 17 at 15:19

Brahadeesh

5,53941956

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

add a comment |

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

Should $p$ be in the set $E$ for the definition $(b)$ ?
– HindShah
Nov 17 at 16:37

@HindShah no, it is not necessary that $p$ should be in the set $E$ in the definition (b). The definition does not assume that $p in E$.
– Brahadeesh
Nov 17 at 16:39

@HindShah without additional assumptions in the statement (b), we just assume that “all points and sets are understood to be elements and subsets of $X$”.
– Brahadeesh
Nov 17 at 16:43

add a comment |

up vote
3
down vote

$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.

answered Nov 17 at 12:48

Henno Brandsma

101k344107

add a comment |

up vote
3
down vote

$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.

answered Nov 17 at 12:48

Henno Brandsma

101k344107

add a comment |

up vote
3
down vote

$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.

answered Nov 17 at 12:48

Henno Brandsma

101k344107

$1$ and $2$ are limit points of $E= (1,2)$ but are not points of $E$. So $E$ is not closed.

answered Nov 17 at 12:48

Henno Brandsma

101k344107

answered Nov 17 at 12:48

Henno Brandsma

101k344107

answered Nov 17 at 12:48

Henno Brandsma

101k344107

answered Nov 17 at 12:48

Henno Brandsma

101k344107

add a comment |

HindShah is a new contributor. Be nice, and check out our Code of Conduct.

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