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If $A$ is a subset of a compact space such that every point of $A$ is an isolated point, is $A$ necessarily...
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Let $A$ be a subset of a compact topological space such that every point of $A$ is an isolated point of $A$. Is $A$ necessarily finite?
general-topology
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
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$begingroup$
Let $A$ be a subset of a compact topological space such that every point of $A$ is an isolated point of $A$. Is $A$ necessarily finite?
general-topology
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
$endgroup$
add a comment |
$begingroup$
Let $A$ be a subset of a compact topological space such that every point of $A$ is an isolated point of $A$. Is $A$ necessarily finite?
general-topology
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
$endgroup$
Let $A$ be a subset of a compact topological space such that every point of $A$ is an isolated point of $A$. Is $A$ necessarily finite?
general-topology
general-topology
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
asked Jan 2 at 16:49
gladimetcampbellsgladimetcampbells
38711
38711
add a comment |
add a comment |
2 Answers
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No. Take for instance the compact space $[0,1]$, and let $A={1/n:ninBbb N_{>0}}$.
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
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No, it can be very large. But if $A$ is also closed, it is compact and then it must be finite.
E.g. $[0,1]^{mathbb{R}}$ is compact in the product topology, but the set $A$ of all functions ${f_t: mathbb{R} to [0,1], t in mathbb{R}}$ defined by $f_t(x) = 0$ if $x neq t$ and $f_t(t)=1$, consists only of isolated points, as $pi_t^{-1}[(frac12,frac32)] cap A = {f_t}$ e.g. So $A$ can be very large, but non-closed.
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
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2 Answers
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2 Answers
2
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$begingroup$
No. Take for instance the compact space $[0,1]$, and let $A={1/n:ninBbb N_{>0}}$.
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
No. Take for instance the compact space $[0,1]$, and let $A={1/n:ninBbb N_{>0}}$.
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
No. Take for instance the compact space $[0,1]$, and let $A={1/n:ninBbb N_{>0}}$.
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
$endgroup$
No. Take for instance the compact space $[0,1]$, and let $A={1/n:ninBbb N_{>0}}$.
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
answered Jan 2 at 16:55
TonyKTonyK
42.1k355134
42.1k355134
add a comment |
add a comment |
$begingroup$
No, it can be very large. But if $A$ is also closed, it is compact and then it must be finite.
E.g. $[0,1]^{mathbb{R}}$ is compact in the product topology, but the set $A$ of all functions ${f_t: mathbb{R} to [0,1], t in mathbb{R}}$ defined by $f_t(x) = 0$ if $x neq t$ and $f_t(t)=1$, consists only of isolated points, as $pi_t^{-1}[(frac12,frac32)] cap A = {f_t}$ e.g. So $A$ can be very large, but non-closed.
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
No, it can be very large. But if $A$ is also closed, it is compact and then it must be finite.
E.g. $[0,1]^{mathbb{R}}$ is compact in the product topology, but the set $A$ of all functions ${f_t: mathbb{R} to [0,1], t in mathbb{R}}$ defined by $f_t(x) = 0$ if $x neq t$ and $f_t(t)=1$, consists only of isolated points, as $pi_t^{-1}[(frac12,frac32)] cap A = {f_t}$ e.g. So $A$ can be very large, but non-closed.
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
add a comment |
$begingroup$
No, it can be very large. But if $A$ is also closed, it is compact and then it must be finite.
E.g. $[0,1]^{mathbb{R}}$ is compact in the product topology, but the set $A$ of all functions ${f_t: mathbb{R} to [0,1], t in mathbb{R}}$ defined by $f_t(x) = 0$ if $x neq t$ and $f_t(t)=1$, consists only of isolated points, as $pi_t^{-1}[(frac12,frac32)] cap A = {f_t}$ e.g. So $A$ can be very large, but non-closed.
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
No, it can be very large. But if $A$ is also closed, it is compact and then it must be finite.
E.g. $[0,1]^{mathbb{R}}$ is compact in the product topology, but the set $A$ of all functions ${f_t: mathbb{R} to [0,1], t in mathbb{R}}$ defined by $f_t(x) = 0$ if $x neq t$ and $f_t(t)=1$, consists only of isolated points, as $pi_t^{-1}[(frac12,frac32)] cap A = {f_t}$ e.g. So $A$ can be very large, but non-closed.
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
answered Jan 2 at 17:51
Henno BrandsmaHenno Brandsma
106k347114
106k347114
add a comment |
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