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
asked 17 hours ago
davidh
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up vote
1
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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
asked 17 hours ago
davidh
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
add a comment |
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
asked 17 hours ago
davidh
605
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
asked 17 hours ago
davidh
605
asked 17 hours ago
davidh
605
asked 17 hours ago
davidh
605
asked 17 hours ago
davidh
605
asked 17 hours ago
davidh
605
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
add a comment |
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
add a comment |
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 ${z:d_2(x,z)>r}.$
answered 17 hours ago
Kavi Rama Murthy
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
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 ${z:d_2(x,z)>r}.$
answered 17 hours ago
Kavi Rama Murthy
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
|
show 4 more comments
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 ${z:d_2(x,z)>r}.$
answered 17 hours ago
Kavi Rama Murthy
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
|
show 4 more comments
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 ${z:d_2(x,z)>r}.$
answered 17 hours ago
Kavi Rama Murthy
43.3k31751
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 ${z:d_2(x,z)>r}.$
answered 17 hours ago
Kavi Rama Murthy
43.3k31751
edited 17 hours ago
answered 17 hours ago
Kavi Rama Murthy
43.3k31751
answered 17 hours ago
Kavi Rama Murthy
43.3k31751
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
|
show 4 more comments
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
|
show 4 more comments
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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