Prove that if a set is nowhere dense iff the complement of the closure of the set is dense.
up vote
4
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I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.
real-analysis general-topology
add a comment |
up vote
4
down vote
favorite
I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.
real-analysis general-topology
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.
real-analysis general-topology
I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.
real-analysis general-topology
real-analysis general-topology
edited Jan 28 '16 at 5:41
Math Wizard
13.2k11036
13.2k11036
asked Jan 28 '16 at 5:31
frank
212
212
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add a comment |
4 Answers
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Let $U subseteq X$ be an open set and let $S subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U cap bar{S}^c neq emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.
add a comment |
up vote
1
down vote
Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.
To prove that $S$ is nowhere dense, any open set $U subseteq cl(S)$ must be empty. Let $U$ be such a set.
The complements are included in the opposite order, that is $cl(S)^{c} subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) subseteq U^{c}$.
But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X subseteq U^{c}$. This is possible only if $U = emptyset$.
Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.
add a comment |
up vote
0
down vote
Let $A^c$ be complement of $A$, $overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $overline{A}=X$.
2). $A$ is nowhere dense iff $(overline{A})^o=varnothing$.
Edit:
We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma
$$
overline{A}=((A^c)^o)^ctag1
$$
By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^osubset A^c$. Thus $A=(A^c)^csubset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.
Now if $A$ be nowhere dense, then $(overline{A})^o=varnothing$. Since by $(1)$ and Involution
$$
overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c=varnothing^c=X
$$
i.e. $(overline{A})^c$ is dense.
Second if $(overline{A})^c$ is dense, then $overline{(overline{A})^c}=X$ and
$$
((overline{A})^o)^c=((((overline{A})^{c})^c)^o)^c=overline{(overline{A})^c}=X
$$
So $(overline{A})^o=varnothing$, i.e $A$ is nowhere dense.
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
|
show 2 more comments
up vote
0
down vote
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
add a comment |
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4 Answers
4
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4 Answers
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up vote
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Let $U subseteq X$ be an open set and let $S subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U cap bar{S}^c neq emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.
add a comment |
up vote
1
down vote
Let $U subseteq X$ be an open set and let $S subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U cap bar{S}^c neq emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $U subseteq X$ be an open set and let $S subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U cap bar{S}^c neq emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.
Let $U subseteq X$ be an open set and let $S subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U cap bar{S}^c neq emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.
answered Jan 28 '16 at 5:41
basket
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1,6771515
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up vote
1
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Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.
To prove that $S$ is nowhere dense, any open set $U subseteq cl(S)$ must be empty. Let $U$ be such a set.
The complements are included in the opposite order, that is $cl(S)^{c} subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) subseteq U^{c}$.
But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X subseteq U^{c}$. This is possible only if $U = emptyset$.
Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.
add a comment |
up vote
1
down vote
Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.
To prove that $S$ is nowhere dense, any open set $U subseteq cl(S)$ must be empty. Let $U$ be such a set.
The complements are included in the opposite order, that is $cl(S)^{c} subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) subseteq U^{c}$.
But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X subseteq U^{c}$. This is possible only if $U = emptyset$.
Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.
To prove that $S$ is nowhere dense, any open set $U subseteq cl(S)$ must be empty. Let $U$ be such a set.
The complements are included in the opposite order, that is $cl(S)^{c} subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) subseteq U^{c}$.
But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X subseteq U^{c}$. This is possible only if $U = emptyset$.
Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.
Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.
To prove that $S$ is nowhere dense, any open set $U subseteq cl(S)$ must be empty. Let $U$ be such a set.
The complements are included in the opposite order, that is $cl(S)^{c} subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) subseteq U^{c}$.
But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X subseteq U^{c}$. This is possible only if $U = emptyset$.
Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.
answered Aug 17 '17 at 8:06
Jan Vysoky
1138
1138
add a comment |
add a comment |
up vote
0
down vote
Let $A^c$ be complement of $A$, $overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $overline{A}=X$.
2). $A$ is nowhere dense iff $(overline{A})^o=varnothing$.
Edit:
We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma
$$
overline{A}=((A^c)^o)^ctag1
$$
By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^osubset A^c$. Thus $A=(A^c)^csubset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.
Now if $A$ be nowhere dense, then $(overline{A})^o=varnothing$. Since by $(1)$ and Involution
$$
overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c=varnothing^c=X
$$
i.e. $(overline{A})^c$ is dense.
Second if $(overline{A})^c$ is dense, then $overline{(overline{A})^c}=X$ and
$$
((overline{A})^o)^c=((((overline{A})^{c})^c)^o)^c=overline{(overline{A})^c}=X
$$
So $(overline{A})^o=varnothing$, i.e $A$ is nowhere dense.
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
|
show 2 more comments
up vote
0
down vote
Let $A^c$ be complement of $A$, $overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $overline{A}=X$.
2). $A$ is nowhere dense iff $(overline{A})^o=varnothing$.
Edit:
We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma
$$
overline{A}=((A^c)^o)^ctag1
$$
By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^osubset A^c$. Thus $A=(A^c)^csubset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.
Now if $A$ be nowhere dense, then $(overline{A})^o=varnothing$. Since by $(1)$ and Involution
$$
overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c=varnothing^c=X
$$
i.e. $(overline{A})^c$ is dense.
Second if $(overline{A})^c$ is dense, then $overline{(overline{A})^c}=X$ and
$$
((overline{A})^o)^c=((((overline{A})^{c})^c)^o)^c=overline{(overline{A})^c}=X
$$
So $(overline{A})^o=varnothing$, i.e $A$ is nowhere dense.
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
|
show 2 more comments
up vote
0
down vote
up vote
0
down vote
Let $A^c$ be complement of $A$, $overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $overline{A}=X$.
2). $A$ is nowhere dense iff $(overline{A})^o=varnothing$.
Edit:
We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma
$$
overline{A}=((A^c)^o)^ctag1
$$
By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^osubset A^c$. Thus $A=(A^c)^csubset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.
Now if $A$ be nowhere dense, then $(overline{A})^o=varnothing$. Since by $(1)$ and Involution
$$
overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c=varnothing^c=X
$$
i.e. $(overline{A})^c$ is dense.
Second if $(overline{A})^c$ is dense, then $overline{(overline{A})^c}=X$ and
$$
((overline{A})^o)^c=((((overline{A})^{c})^c)^o)^c=overline{(overline{A})^c}=X
$$
So $(overline{A})^o=varnothing$, i.e $A$ is nowhere dense.
Let $A^c$ be complement of $A$, $overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.
We define:
1). $A$ is dense iff $overline{A}=X$.
2). $A$ is nowhere dense iff $(overline{A})^o=varnothing$.
Edit:
We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma
$$
overline{A}=((A^c)^o)^ctag1
$$
By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^osubset A^c$. Thus $A=(A^c)^csubset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.
Now if $A$ be nowhere dense, then $(overline{A})^o=varnothing$. Since by $(1)$ and Involution
$$
overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c=varnothing^c=X
$$
i.e. $(overline{A})^c$ is dense.
Second if $(overline{A})^c$ is dense, then $overline{(overline{A})^c}=X$ and
$$
((overline{A})^o)^c=((((overline{A})^{c})^c)^o)^c=overline{(overline{A})^c}=X
$$
So $(overline{A})^o=varnothing$, i.e $A$ is nowhere dense.
edited Jan 30 '16 at 4:46
answered Jan 28 '16 at 6:04
Math Wizard
13.2k11036
13.2k11036
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
|
show 2 more comments
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
I am not sure how you simplified the initial expressions in the two steps.
– frank
Jan 28 '16 at 6:28
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $overline{A}=(((A)^c)^o)^c$.
– Math Wizard
Jan 28 '16 at 6:29
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps.
– frank
Jan 28 '16 at 6:34
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
By $overline{A}=(((A)^c)^o)^c$, we have $overline{(overline{A})^c}=((((overline{A})^{c})^c)^o)^c$. By involution, $((overline{A})^{c})^c=overline{A}$. So $((((overline{A})^{c})^c)^o)^c=((overline{A})^o)^c$
– Math Wizard
Jan 28 '16 at 6:37
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
Can you explain why the first fact is true?
– frank
Jan 28 '16 at 6:39
|
show 2 more comments
up vote
0
down vote
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
add a comment |
up vote
0
down vote
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
add a comment |
up vote
0
down vote
up vote
0
down vote
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
Assume that complement of (clA) is dense.
Then cl[complement of (clA)] = X.
So complement of (int clA) = X.
Thus int clA is empty.
Hence A is nowhere dense.
answered Nov 23 at 2:43
P.SUNDARAM
211
211
add a comment |
add a comment |
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