Let $mathbb{R}^n$ be endowed with the Euclidean metric $d_2$. Let S be a nonempty subset of $mathbb{R}^n$











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1) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ closed in S?



2) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ contained in the closure of ${yin S: d_2(x,y) > r}$ in S?



my intuition for both is yes...
For 1). I think I need to prove the complement of set in S is open. However, I don't know how to quantify the argument.






real-analysis







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    A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
    – GNUSupporter 8964民主女神 地下教會
    17 hours ago















up vote
1
down vote

favorite












1) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ closed in S?



2) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ contained in the closure of ${yin S: d_2(x,y) > r}$ in S?



my intuition for both is yes...
For 1). I think I need to prove the complement of set in S is open. However, I don't know how to quantify the argument.






real-analysis







share|cite|improve this question


















  • 1




    A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
    – GNUSupporter 8964民主女神 地下教會
    17 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











1) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ closed in S?



2) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ contained in the closure of ${yin S: d_2(x,y) > r}$ in S?



my intuition for both is yes...
For 1). I think I need to prove the complement of set in S is open. However, I don't know how to quantify the argument.






real-analysis







share|cite|improve this question













1) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ closed in S?



2) Given $xin S$, is the set ${yin S: d_2(x,y) geq r}$ contained in the closure of ${yin S: d_2(x,y) > r}$ in S?



my intuition for both is yes...
For 1). I think I need to prove the complement of set in S is open. However, I don't know how to quantify the argument.






real-analysis




real-analysis






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asked 17 hours ago









davidh

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605








  • 1




    A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
    – GNUSupporter 8964民主女神 地下教會
    17 hours ago














  • 1




    A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
    – GNUSupporter 8964民主女神 地下教會
    17 hours ago








1




1




A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
– GNUSupporter 8964民主女神 地下教會
17 hours ago




A metric is by definition continuous in the metric space. The preimage of an open set by a continuous function is open. This should solve most of the problems.
– GNUSupporter 8964民主女神 地下教會
17 hours ago










1 Answer
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Let $S'$ be the set in the statement. Let ${x_k} subset S'$ and $x_k to x'$. Then $r leq d_2(x_k,x)leq d_2(x',x)+d_2(x_k,x')$. Let $k to infty $ to see that $x in S'$. hence $S'$ is closed.



Now let $d_2(x,y) geq r$. Let $x_k=y+frac 1 k (y-x)$. Then $d_2(x_k,y)=d_2(0,frac 1 k (y-x)) to 0$ as $ k to infty$ . Also, $$d_2(x_k,x)=d_2(y+frac 1 k (y-x),x)$$ $$=d_2((y-x+frac 1 k (y-x),0)$$ $$ =(1+frac 1 k) d_2(x,y) >d_2(x,y) geq r.$$ Hence $y$ is the limit of a sequence from ${z:d_2(x,z)>r}.$






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  • i don't think I get the first part tho, is the sequence's limit x the same x in the question?
    – davidh
    17 hours ago










  • @davidh Sorry, that was a mistake. I have corrected it.
    – Kavi Rama Murthy
    17 hours ago










  • I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
    – davidh
    17 hours ago












  • @davidh Corrected that also. Hope I have fixed all the errors.
    – Kavi Rama Murthy
    17 hours ago










  • Thanks. Could you explain how you got the (1+1/k)^n step?
    – davidh
    17 hours ago











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1 Answer
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1 Answer
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active

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active

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up vote
1
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accepted










Let $S'$ be the set in the statement. Let ${x_k} subset S'$ and $x_k to x'$. Then $r leq d_2(x_k,x)leq d_2(x',x)+d_2(x_k,x')$. Let $k to infty $ to see that $x in S'$. hence $S'$ is closed.



Now let $d_2(x,y) geq r$. Let $x_k=y+frac 1 k (y-x)$. Then $d_2(x_k,y)=d_2(0,frac 1 k (y-x)) to 0$ as $ k to infty$ . Also, $$d_2(x_k,x)=d_2(y+frac 1 k (y-x),x)$$ $$=d_2((y-x+frac 1 k (y-x),0)$$ $$ =(1+frac 1 k) d_2(x,y) >d_2(x,y) geq r.$$ Hence $y$ is the limit of a sequence from ${z:d_2(x,z)>r}.$






share|cite|improve this answer























  • i don't think I get the first part tho, is the sequence's limit x the same x in the question?
    – davidh
    17 hours ago










  • @davidh Sorry, that was a mistake. I have corrected it.
    – Kavi Rama Murthy
    17 hours ago










  • I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
    – davidh
    17 hours ago












  • @davidh Corrected that also. Hope I have fixed all the errors.
    – Kavi Rama Murthy
    17 hours ago










  • Thanks. Could you explain how you got the (1+1/k)^n step?
    – davidh
    17 hours ago















up vote
1
down vote



accepted










Let $S'$ be the set in the statement. Let ${x_k} subset S'$ and $x_k to x'$. Then $r leq d_2(x_k,x)leq d_2(x',x)+d_2(x_k,x')$. Let $k to infty $ to see that $x in S'$. hence $S'$ is closed.



Now let $d_2(x,y) geq r$. Let $x_k=y+frac 1 k (y-x)$. Then $d_2(x_k,y)=d_2(0,frac 1 k (y-x)) to 0$ as $ k to infty$ . Also, $$d_2(x_k,x)=d_2(y+frac 1 k (y-x),x)$$ $$=d_2((y-x+frac 1 k (y-x),0)$$ $$ =(1+frac 1 k) d_2(x,y) >d_2(x,y) geq r.$$ Hence $y$ is the limit of a sequence from ${z:d_2(x,z)>r}.$






share|cite|improve this answer























  • i don't think I get the first part tho, is the sequence's limit x the same x in the question?
    – davidh
    17 hours ago










  • @davidh Sorry, that was a mistake. I have corrected it.
    – Kavi Rama Murthy
    17 hours ago










  • I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
    – davidh
    17 hours ago












  • @davidh Corrected that also. Hope I have fixed all the errors.
    – Kavi Rama Murthy
    17 hours ago










  • Thanks. Could you explain how you got the (1+1/k)^n step?
    – davidh
    17 hours ago













up vote
1
down vote



accepted







up vote
1
down vote



accepted






Let $S'$ be the set in the statement. Let ${x_k} subset S'$ and $x_k to x'$. Then $r leq d_2(x_k,x)leq d_2(x',x)+d_2(x_k,x')$. Let $k to infty $ to see that $x in S'$. hence $S'$ is closed.



Now let $d_2(x,y) geq r$. Let $x_k=y+frac 1 k (y-x)$. Then $d_2(x_k,y)=d_2(0,frac 1 k (y-x)) to 0$ as $ k to infty$ . Also, $$d_2(x_k,x)=d_2(y+frac 1 k (y-x),x)$$ $$=d_2((y-x+frac 1 k (y-x),0)$$ $$ =(1+frac 1 k) d_2(x,y) >d_2(x,y) geq r.$$ Hence $y$ is the limit of a sequence from ${z:d_2(x,z)>r}.$






share|cite|improve this answer














Let $S'$ be the set in the statement. Let ${x_k} subset S'$ and $x_k to x'$. Then $r leq d_2(x_k,x)leq d_2(x',x)+d_2(x_k,x')$. Let $k to infty $ to see that $x in S'$. hence $S'$ is closed.



Now let $d_2(x,y) geq r$. Let $x_k=y+frac 1 k (y-x)$. Then $d_2(x_k,y)=d_2(0,frac 1 k (y-x)) to 0$ as $ k to infty$ . Also, $$d_2(x_k,x)=d_2(y+frac 1 k (y-x),x)$$ $$=d_2((y-x+frac 1 k (y-x),0)$$ $$ =(1+frac 1 k) d_2(x,y) >d_2(x,y) geq r.$$ Hence $y$ is the limit of a sequence from ${z:d_2(x,z)>r}.$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 17 hours ago

























answered 17 hours ago









Kavi Rama Murthy

43.3k31751




43.3k31751












  • i don't think I get the first part tho, is the sequence's limit x the same x in the question?
    – davidh
    17 hours ago










  • @davidh Sorry, that was a mistake. I have corrected it.
    – Kavi Rama Murthy
    17 hours ago










  • I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
    – davidh
    17 hours ago












  • @davidh Corrected that also. Hope I have fixed all the errors.
    – Kavi Rama Murthy
    17 hours ago










  • Thanks. Could you explain how you got the (1+1/k)^n step?
    – davidh
    17 hours ago


















  • i don't think I get the first part tho, is the sequence's limit x the same x in the question?
    – davidh
    17 hours ago










  • @davidh Sorry, that was a mistake. I have corrected it.
    – Kavi Rama Murthy
    17 hours ago










  • I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
    – davidh
    17 hours ago












  • @davidh Corrected that also. Hope I have fixed all the errors.
    – Kavi Rama Murthy
    17 hours ago










  • Thanks. Could you explain how you got the (1+1/k)^n step?
    – davidh
    17 hours ago
















i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
17 hours ago




i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
17 hours ago












@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
17 hours ago




@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
17 hours ago












I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
17 hours ago






I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
17 hours ago














@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
17 hours ago




@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
17 hours ago












Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
17 hours ago




Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
17 hours ago


















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