Mixing
Difference between sequentially compact spaces and closed sets in $mathbb{C}$
$begingroup$
According to my notes
$Ksubsetmathbb{C}$ sequentially compact $iff$ Every sequence in K has a subsequence which converges to a Point in K
$Asubseteqmathbb{C}$ closed set $iff ((a_n)_{ninmathbb{N}}$ convergent sequence in A $Rightarrow lim_{nrightarrow infty}a_nin A$
I am wondering why the definitions, save for $mathbb{C}$ are not equivalent
If $Aneqmathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.
general-topology compactness
edited Jan 13 at 13:47
Paul Frost
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
$endgroup$
add a comment |
$begingroup$
According to my notes
$Ksubsetmathbb{C}$ sequentially compact $iff$ Every sequence in K has a subsequence which converges to a Point in K
$Asubseteqmathbb{C}$ closed set $iff ((a_n)_{ninmathbb{N}}$ convergent sequence in A $Rightarrow lim_{nrightarrow infty}a_nin A$
I am wondering why the definitions, save for $mathbb{C}$ are not equivalent
If $Aneqmathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.
general-topology compactness
edited Jan 13 at 13:47
Paul Frost
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
$endgroup$
$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05
add a comment |
$begingroup$
According to my notes
$Ksubsetmathbb{C}$ sequentially compact $iff$ Every sequence in K has a subsequence which converges to a Point in K
$Asubseteqmathbb{C}$ closed set $iff ((a_n)_{ninmathbb{N}}$ convergent sequence in A $Rightarrow lim_{nrightarrow infty}a_nin A$
I am wondering why the definitions, save for $mathbb{C}$ are not equivalent
If $Aneqmathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.
general-topology compactness
edited Jan 13 at 13:47
Paul Frost
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
$endgroup$
According to my notes
$Ksubsetmathbb{C}$ sequentially compact $iff$ Every sequence in K has a subsequence which converges to a Point in K
$Asubseteqmathbb{C}$ closed set $iff ((a_n)_{ninmathbb{N}}$ convergent sequence in A $Rightarrow lim_{nrightarrow infty}a_nin A$
I am wondering why the definitions, save for $mathbb{C}$ are not equivalent
If $Aneqmathbb{C}$ and $A$ closed set, why does $A$ not have to be necessarily compact? Because the opposite is true. If I take a convergent sequence of a sequentially compact subset $K$ then the convergencepoint has to be by Definition in $K$.
general-topology compactness
general-topology compactness
edited Jan 13 at 13:47
Paul Frost
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
edited Jan 13 at 13:47
Paul Frost
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
edited Jan 13 at 13:47
Paul Frost
10.7k3934
edited Jan 13 at 13:47
Paul Frost
10.7k3934
edited Jan 13 at 13:47
Paul Frost
10.7k3934
10.7k3934
asked Jan 13 at 12:50
RM777RM777
37612
asked Jan 13 at 12:50
RM777RM777
37612
asked Jan 13 at 12:50
RM777RM777
37612
37612
$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05
add a comment |
$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05
$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05
$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take $A={zinmathbb{C},|,lvert zrvertgeqslant1}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Take $A={zinmathbb{C},|,lvert zrvertgeqslant1}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
add a comment |
$begingroup$
Take $A={zinmathbb{C},|,lvert zrvertgeqslant1}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
add a comment |
$begingroup$
Take $A={zinmathbb{C},|,lvert zrvertgeqslant1}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
Take $A={zinmathbb{C},|,lvert zrvertgeqslant1}$, for instance. It is closed. However, it is not sequentially compact because, for instance, the sequence $1,2,3,ldots$ has no convergente subsequence. All you can deduce about sequences of elements of $A$ from the fact that $A$ is closed is that if one such sequence converges, then its limit belongs to $A$.
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
answered Jan 13 at 12:55
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
add a comment |
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
$begingroup$
You said I should take the sequence $1,2,3,...$ but $2$ and $3$ are not Elements of $A$.
$endgroup$
– RM777
Jan 13 at 13:02
1
1
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
$begingroup$
Why? After all, $lvert2rvert=2geqslant1$ and $lvert3rvert=3geqslant1$.
$endgroup$
– José Carlos Santos
Jan 13 at 13:03
add a comment |
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$begingroup$
sequentially compact (in $mathbb{C}$) is equivalent to "closed and bounded" by Heine-Borel. So unbounded closed sets are counterexamples.
$endgroup$
– Henno Brandsma
Jan 13 at 15:05