Limit points of infinite subsets of closed sets
$begingroup$
Is the following statement true or false?
If $F$ is an infinite subset of a closed set $E$, then $F$ has a limit point in $E$?
The original one is: if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$. This statement is proved in Rudin's Principles of Mathematical Analysis. However, I just use the closing property of $K$ in my proof, so I'm not sure of the statemet above.
general-topology
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
$endgroup$
add a comment |
$begingroup$
Is the following statement true or false?
If $F$ is an infinite subset of a closed set $E$, then $F$ has a limit point in $E$?
The original one is: if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$. This statement is proved in Rudin's Principles of Mathematical Analysis. However, I just use the closing property of $K$ in my proof, so I'm not sure of the statemet above.
general-topology
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
$endgroup$
2
$begingroup$
What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17
add a comment |
$begingroup$
Is the following statement true or false?
If $F$ is an infinite subset of a closed set $E$, then $F$ has a limit point in $E$?
The original one is: if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$. This statement is proved in Rudin's Principles of Mathematical Analysis. However, I just use the closing property of $K$ in my proof, so I'm not sure of the statemet above.
general-topology
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
$endgroup$
Is the following statement true or false?
If $F$ is an infinite subset of a closed set $E$, then $F$ has a limit point in $E$?
The original one is: if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$. This statement is proved in Rudin's Principles of Mathematical Analysis. However, I just use the closing property of $K$ in my proof, so I'm not sure of the statemet above.
general-topology
general-topology
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
edited Aug 21 '14 at 10:12
Meta-мета-μετα-meta-мета-μετα
4,724927
4,724927
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
asked Aug 21 '14 at 5:15
Tien Kha PhamTien Kha Pham
1,717922
1,717922
2
$begingroup$
What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17
add a comment |
2
$begingroup$
What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17
2
2
$begingroup$
What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17
$begingroup$
What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17
add a comment |
1 Answer
1
active
oldest
votes
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Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $mathbb{R}$ (in the space $mathbb{R}$), and take $mathbb{Z}$.
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
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add a comment |
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$begingroup$
Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $mathbb{R}$ (in the space $mathbb{R}$), and take $mathbb{Z}$.
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
$endgroup$
add a comment |
$begingroup$
Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $mathbb{R}$ (in the space $mathbb{R}$), and take $mathbb{Z}$.
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
$endgroup$
add a comment |
$begingroup$
Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $mathbb{R}$ (in the space $mathbb{R}$), and take $mathbb{Z}$.
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
$endgroup$
Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $mathbb{R}$ (in the space $mathbb{R}$), and take $mathbb{Z}$.
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
answered Aug 21 '14 at 5:55
Alex R.Alex R.
25.2k12452
25.2k12452
add a comment |
add a comment |
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What about if you consider $mathbb N subset [0, infty)$, where all are considered as subspaces of $mathbb R$ with the usual topology?
$endgroup$
– vociferous_rutabaga
Aug 21 '14 at 5:17