Mixing
$X=(-infty,+infty]$, $T_>:={(-a,+infty]:a in [-infty, +infty]}$. Is $X$ compact?
Let $X=(-infty,+infty]$ and $T_>:={(-a,+infty]:a in [-infty, +infty]}$.
$T_>$ is obviously a topology. How can I reason that $X$ is not compact? That is there exists an open cover of $X$ which has no finite subcover.
I thought about $cup_{n in mathbb{N}}(-n,+infty]$ which is an open cover of $X$ but for every $N in mathbb{N}$ $cup_{n=1}^{N}(-n, +infty)$ does not cover $(-infty,N]$ so $cup_{n in mathbb{N}}(-n,+infty]$ has no finite subcover.
general-topology proof-verification
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
asked Nov 20 '18 at 21:23
conrad
757
add a comment |
Let $X=(-infty,+infty]$ and $T_>:={(-a,+infty]:a in [-infty, +infty]}$.
$T_>$ is obviously a topology. How can I reason that $X$ is not compact? That is there exists an open cover of $X$ which has no finite subcover.
I thought about $cup_{n in mathbb{N}}(-n,+infty]$ which is an open cover of $X$ but for every $N in mathbb{N}$ $cup_{n=1}^{N}(-n, +infty)$ does not cover $(-infty,N]$ so $cup_{n in mathbb{N}}(-n,+infty]$ has no finite subcover.
general-topology proof-verification
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
asked Nov 20 '18 at 21:23
conrad
757
add a comment |
Let $X=(-infty,+infty]$ and $T_>:={(-a,+infty]:a in [-infty, +infty]}$.
$T_>$ is obviously a topology. How can I reason that $X$ is not compact? That is there exists an open cover of $X$ which has no finite subcover.
I thought about $cup_{n in mathbb{N}}(-n,+infty]$ which is an open cover of $X$ but for every $N in mathbb{N}$ $cup_{n=1}^{N}(-n, +infty)$ does not cover $(-infty,N]$ so $cup_{n in mathbb{N}}(-n,+infty]$ has no finite subcover.
general-topology proof-verification
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
asked Nov 20 '18 at 21:23
conrad
757
Let $X=(-infty,+infty]$ and $T_>:={(-a,+infty]:a in [-infty, +infty]}$.
$T_>$ is obviously a topology. How can I reason that $X$ is not compact? That is there exists an open cover of $X$ which has no finite subcover.
I thought about $cup_{n in mathbb{N}}(-n,+infty]$ which is an open cover of $X$ but for every $N in mathbb{N}$ $cup_{n=1}^{N}(-n, +infty)$ does not cover $(-infty,N]$ so $cup_{n in mathbb{N}}(-n,+infty]$ has no finite subcover.
general-topology proof-verification
general-topology proof-verification
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
asked Nov 20 '18 at 21:23
conrad
757
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
asked Nov 20 '18 at 21:23
conrad
757
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
edited Nov 20 '18 at 21:30
Henno Brandsma
105k347114
105k347114
asked Nov 20 '18 at 21:23
conrad
757
asked Nov 20 '18 at 21:23
conrad
757
asked Nov 20 '18 at 21:23
conrad
757
757
add a comment |
add a comment |
1 Answer
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Yes, the cover $(-n, +infty]$ is a nested cover (the sets grow larger with increasing $n$) and so any finite subset of them has as their union the one with the maximal $n$, and so a finite subset never covers all of $X$.
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
Yes, the cover $(-n, +infty]$ is a nested cover (the sets grow larger with increasing $n$) and so any finite subset of them has as their union the one with the maximal $n$, and so a finite subset never covers all of $X$.
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
add a comment |
Yes, the cover $(-n, +infty]$ is a nested cover (the sets grow larger with increasing $n$) and so any finite subset of them has as their union the one with the maximal $n$, and so a finite subset never covers all of $X$.
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
add a comment |
Yes, the cover $(-n, +infty]$ is a nested cover (the sets grow larger with increasing $n$) and so any finite subset of them has as their union the one with the maximal $n$, and so a finite subset never covers all of $X$.
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
Yes, the cover $(-n, +infty]$ is a nested cover (the sets grow larger with increasing $n$) and so any finite subset of them has as their union the one with the maximal $n$, and so a finite subset never covers all of $X$.
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:29
Henno Brandsma
105k347114
105k347114
add a comment |
add a comment |
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