Prove that if a set is nowhere dense iff the complement of the closure of the set is dense.











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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.










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    up vote
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    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.










    share|cite|improve this question


























      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.










      share|cite|improve this question















      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






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      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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          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.






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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.






            share|cite|improve this answer




























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              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.






              share|cite|improve this answer























              • 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


















              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.






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                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.






                share|cite|improve this answer

























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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.






                  share|cite|improve this answer























                    up vote
                    1
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                    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.






                    share|cite|improve this answer












                    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.







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                    answered Jan 28 '16 at 5:41









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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.






                        share|cite|improve this answer

























                          up vote
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                          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.






                          share|cite|improve this answer























                            up vote
                            1
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                            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.






                            share|cite|improve this answer












                            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.







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                            answered Aug 17 '17 at 8:06









                            Jan Vysoky

                            1138




                            1138






















                                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.






                                share|cite|improve this answer























                                • 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















                                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.






                                share|cite|improve this answer























                                • 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













                                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.






                                share|cite|improve this answer














                                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.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                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


















                                • 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










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                                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.






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                                  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.






                                  share|cite|improve this answer























                                    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.






                                    share|cite|improve this answer












                                    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.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 23 at 2:43









                                    P.SUNDARAM

                                    211




                                    211






























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