Let $mathbb{R}^n$ be endowed with the Euclidean metric $d_2$. Let S be a nonempty subset of $mathbb{R}^n$
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
add a comment |
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
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
Nov 30 at 9:51
add a comment |
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
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 Nov 30 at 9:44
davidh
695
695
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
Nov 30 at 9:51
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
Nov 30 at 9:51
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
Nov 30 at 9:51
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
Nov 30 at 9:51
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 ${z:d_2(x,z)>r}.$
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
|
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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 ${z:d_2(x,z)>r}.$
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
|
show 4 more comments
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 ${z:d_2(x,z)>r}.$
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
|
show 4 more comments
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 ${z:d_2(x,z)>r}.$
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 ${z:d_2(x,z)>r}.$
edited Nov 30 at 10:18
answered Nov 30 at 9:54
Kavi Rama Murthy
48.6k31854
48.6k31854
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
|
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
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
i don't think I get the first part tho, is the sequence's limit x the same x in the question?
– davidh
Nov 30 at 10:04
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
@davidh Sorry, that was a mistake. I have corrected it.
– Kavi Rama Murthy
Nov 30 at 10:06
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
I see. however shouldn't the first inequality be $r <= d_2(x_k, x)$?
– davidh
Nov 30 at 10:07
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
@davidh Corrected that also. Hope I have fixed all the errors.
– Kavi Rama Murthy
Nov 30 at 10:09
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
Thanks. Could you explain how you got the (1+1/k)^n step?
– davidh
Nov 30 at 10:11
|
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
Nov 30 at 9:51