Is this topology related problem wrong?
Clash Royale CLAN TAG#URR8PPP
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One of my friend sent me this question. I think it is a wrong question
Why?
When we define topology on $X$, we talk about set of all subsets of $X$. Here $(1,2018]$ is not even a subset of $X$, so we can not say anything about whether it is open or closed in $X$.
general-topology
asked Sep 2 at 12:12
StammeringMathematician
53115
add a comment |Â
up vote
2
down vote
favorite
One of my friend sent me this question. I think it is a wrong question
Why?
When we define topology on $X$, we talk about set of all subsets of $X$. Here $(1,2018]$ is not even a subset of $X$, so we can not say anything about whether it is open or closed in $X$.
general-topology
asked Sep 2 at 12:12
StammeringMathematician
53115
Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
One of my friend sent me this question. I think it is a wrong question
Why?
When we define topology on $X$, we talk about set of all subsets of $X$. Here $(1,2018]$ is not even a subset of $X$, so we can not say anything about whether it is open or closed in $X$.
general-topology
asked Sep 2 at 12:12
StammeringMathematician
53115
One of my friend sent me this question. I think it is a wrong question
Why?
When we define topology on $X$, we talk about set of all subsets of $X$. Here $(1,2018]$ is not even a subset of $X$, so we can not say anything about whether it is open or closed in $X$.
general-topology
general-topology
asked Sep 2 at 12:12
StammeringMathematician
53115
asked Sep 2 at 12:12
StammeringMathematician
53115
asked Sep 2 at 12:12
StammeringMathematician
53115
asked Sep 2 at 12:12
StammeringMathematician
53115
asked Sep 2 at 12:12
StammeringMathematician
53115
53115
Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14
add a comment |Â
Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14
Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14
Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
$1$ and $2$ are indeed meaningless because $(1,2018]$ is not even a subset of $X$.
$3$ is not true because $1$ is not an interior point of it.
$4$ is also meaningless: $0 notin X$ so it is not even a candidate to be a limit point of any subset of $X$. Its negation is really voidly true:
$0$ is a limit point of $A$ iff for all open sets of $X$, if $0 in O$ then $O cap Asetminusx neq emptyset$. But we never have $0 in O$ so the implication is always true and so the statement on the right hand side is true and $0$ is a limit point of $A$ (whatever $A$ is). So we could defend the negation of 4 to be true, so likewise that 4 is false.
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$1$ and $2$ are indeed meaningless because $(1,2018]$ is not even a subset of $X$.
$3$ is not true because $1$ is not an interior point of it.
$4$ is also meaningless: $0 notin X$ so it is not even a candidate to be a limit point of any subset of $X$. Its negation is really voidly true:
$0$ is a limit point of $A$ iff for all open sets of $X$, if $0 in O$ then $O cap Asetminusx neq emptyset$. But we never have $0 in O$ so the implication is always true and so the statement on the right hand side is true and $0$ is a limit point of $A$ (whatever $A$ is). So we could defend the negation of 4 to be true, so likewise that 4 is false.
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
add a comment |Â
up vote
1
down vote
accepted
$1$ and $2$ are indeed meaningless because $(1,2018]$ is not even a subset of $X$.
$3$ is not true because $1$ is not an interior point of it.
$4$ is also meaningless: $0 notin X$ so it is not even a candidate to be a limit point of any subset of $X$. Its negation is really voidly true:
$0$ is a limit point of $A$ iff for all open sets of $X$, if $0 in O$ then $O cap Asetminusx neq emptyset$. But we never have $0 in O$ so the implication is always true and so the statement on the right hand side is true and $0$ is a limit point of $A$ (whatever $A$ is). So we could defend the negation of 4 to be true, so likewise that 4 is false.
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$1$ and $2$ are indeed meaningless because $(1,2018]$ is not even a subset of $X$.
$3$ is not true because $1$ is not an interior point of it.
$4$ is also meaningless: $0 notin X$ so it is not even a candidate to be a limit point of any subset of $X$. Its negation is really voidly true:
$0$ is a limit point of $A$ iff for all open sets of $X$, if $0 in O$ then $O cap Asetminusx neq emptyset$. But we never have $0 in O$ so the implication is always true and so the statement on the right hand side is true and $0$ is a limit point of $A$ (whatever $A$ is). So we could defend the negation of 4 to be true, so likewise that 4 is false.
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
$1$ and $2$ are indeed meaningless because $(1,2018]$ is not even a subset of $X$.
$3$ is not true because $1$ is not an interior point of it.
$4$ is also meaningless: $0 notin X$ so it is not even a candidate to be a limit point of any subset of $X$. Its negation is really voidly true:
$0$ is a limit point of $A$ iff for all open sets of $X$, if $0 in O$ then $O cap Asetminusx neq emptyset$. But we never have $0 in O$ so the implication is always true and so the statement on the right hand side is true and $0$ is a limit point of $A$ (whatever $A$ is). So we could defend the negation of 4 to be true, so likewise that 4 is false.
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
answered Sep 2 at 12:32
Henno Brandsma
93.4k342101
93.4k342101
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
add a comment |Â
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
Thanks for the clarification for (4) point. I did not have any solid reasoning except the fact that $0$ does not belong to $X$. I think it is example of a non complete metric space as $1/n$ is a cauchy sequence but limit does not lie in the space. Am I right?
– StammeringMathematician
Sep 2 at 12:37
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
@MathamanTopologius Sure, it's a non-complete metric space.
– Henno Brandsma
Sep 2 at 12:51
add a comment |Â
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Yes, $2018$ is not in $X$. So $#2$ is vacuously true and $#1$ is false.
– lulu
Sep 2 at 12:14