Does $C[0, 1]$ have uncountable disjoint open sets?covering positively disjoint sets with finite ballsProve that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric.Prove that two closed and disjoint sets, have open disjoint super-sets with dist metric .Show that exists $U$ and $V$ such that $xin V$, $Ksubset U$ and $Ucap V=emptyset$Non-empty intersection of decreasing open setsThere exist disjoint open sets $U_a ni a$ and $U_b ni b$ such that $overlineU_asubseteq U, overlineU_bsubseteq U$Prove that if $A,B$ are closed then, $ exists;U,V$ open sets such that $Ucap V= emptyset$Prove that if $A,B$ are closed and disjoint in a metric space, then they are contained in disjoint neighborhoodsopen singleton metric space
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Does $C[0, 1]$ have uncountable disjoint open sets?
covering positively disjoint sets with finite ballsProve that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric.Prove that two closed and disjoint sets, have open disjoint super-sets with dist metric .Show that exists $U$ and $V$ such that $xin V$, $Ksubset U$ and $Ucap V=emptyset$Non-empty intersection of decreasing open setsThere exist disjoint open sets $U_a ni a$ and $U_b ni b$ such that $overlineU_asubseteq U, overlineU_bsubseteq U$Prove that if $A,B$ are closed then, $ exists;U,V$ open sets such that $Ucap V= emptyset$Prove that if $A,B$ are closed and disjoint in a metric space, then they are contained in disjoint neighborhoodsopen singleton metric space
$begingroup$
Let $(X, d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = x_jin Xmid jin J$ with the property that $$d(x_j, x_k) = 1$$for all $jneq k$, $j, k in J$.
Which of the following statements are true?
(a) If such a set exists in $X$, then there exist open sets $U_j$ in $X$ such that $U_j cap U_k = emptyset$, for all $j neq k$.
(b) There exists such a set $S$ in $C[0, 1]$ with $J$ being uncountable.
My attempt:
Option a is correct. Consider the sets $U_j= B_frac13(x_j)$ which are open (singletons are open sets in discrete metric). Clearly $U_j cap U_k=emptyset$ as $d(x_j, x_k)=1$.
How to disprove option b? Is the explanation correct for option a?
metric-spaces
edited Mar 14 at 10:01
Saad
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
$endgroup$
add a comment |
$begingroup$
Let $(X, d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = x_jin Xmid jin J$ with the property that $$d(x_j, x_k) = 1$$for all $jneq k$, $j, k in J$.
Which of the following statements are true?
(a) If such a set exists in $X$, then there exist open sets $U_j$ in $X$ such that $U_j cap U_k = emptyset$, for all $j neq k$.
(b) There exists such a set $S$ in $C[0, 1]$ with $J$ being uncountable.
My attempt:
Option a is correct. Consider the sets $U_j= B_frac13(x_j)$ which are open (singletons are open sets in discrete metric). Clearly $U_j cap U_k=emptyset$ as $d(x_j, x_k)=1$.
How to disprove option b? Is the explanation correct for option a?
metric-spaces
edited Mar 14 at 10:01
Saad
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
$endgroup$
$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
1
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47
add a comment |
$begingroup$
Let $(X, d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = x_jin Xmid jin J$ with the property that $$d(x_j, x_k) = 1$$for all $jneq k$, $j, k in J$.
Which of the following statements are true?
(a) If such a set exists in $X$, then there exist open sets $U_j$ in $X$ such that $U_j cap U_k = emptyset$, for all $j neq k$.
(b) There exists such a set $S$ in $C[0, 1]$ with $J$ being uncountable.
My attempt:
Option a is correct. Consider the sets $U_j= B_frac13(x_j)$ which are open (singletons are open sets in discrete metric). Clearly $U_j cap U_k=emptyset$ as $d(x_j, x_k)=1$.
How to disprove option b? Is the explanation correct for option a?
metric-spaces
edited Mar 14 at 10:01
Saad
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
$endgroup$
Let $(X, d)$ be a metric space. Let $J$ be an indexing set. Consider a set of the form $S = x_jin Xmid jin J$ with the property that $$d(x_j, x_k) = 1$$for all $jneq k$, $j, k in J$.
Which of the following statements are true?
(a) If such a set exists in $X$, then there exist open sets $U_j$ in $X$ such that $U_j cap U_k = emptyset$, for all $j neq k$.
(b) There exists such a set $S$ in $C[0, 1]$ with $J$ being uncountable.
My attempt:
Option a is correct. Consider the sets $U_j= B_frac13(x_j)$ which are open (singletons are open sets in discrete metric). Clearly $U_j cap U_k=emptyset$ as $d(x_j, x_k)=1$.
How to disprove option b? Is the explanation correct for option a?
metric-spaces
metric-spaces
edited Mar 14 at 10:01
Saad
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
edited Mar 14 at 10:01
Saad
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
edited Mar 14 at 10:01
Saad
20.1k92352
edited Mar 14 at 10:01
Saad
20.1k92352
edited Mar 14 at 10:01
Saad
20.1k92352
20.1k92352
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
asked Mar 14 at 6:56
Mittal GMittal G
1,379516
1,379516
$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
1
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47
add a comment |
$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
1
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47
$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
1
1
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $f_n$ be dense and for each $j in J$ pick $k_j$ such that $d(f_k_j, x_j) <frac 1 2$. Now verify that $jin J to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
add a comment |
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$begingroup$
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $f_n$ be dense and for each $j in J$ pick $k_j$ such that $d(f_k_j, x_j) <frac 1 2$. Now verify that $jin J to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
add a comment |
$begingroup$
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $f_n$ be dense and for each $j in J$ pick $k_j$ such that $d(f_k_j, x_j) <frac 1 2$. Now verify that $jin J to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
add a comment |
$begingroup$
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $f_n$ be dense and for each $j in J$ pick $k_j$ such that $d(f_k_j, x_j) <frac 1 2$. Now verify that $jin J to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
$endgroup$
$C [0,1]$ (with the usual sup metric) is separable. Polynomials with rational coefficients form a countable dense set. Let $f_n$ be dense and for each $j in J$ pick $k_j$ such that $d(f_k_j, x_j) <frac 1 2$. Now verify that $jin J to k_j$ is a one to one map from $j$ into the set of positive integers. Hence $J$ is necessarily countable.
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
answered Mar 14 at 7:52
Kavi Rama MurthyKavi Rama Murthy
68.6k53169
68.6k53169
add a comment |
add a comment |
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$begingroup$
@Saucy O'Path Sorry don't get it.
$endgroup$
– Mittal G
Mar 14 at 6:59
$begingroup$
No separable metric space can have uncounatbly many points at distance $1$ from each other.
$endgroup$
– Kavi Rama Murthy
Mar 14 at 8:00
1
$begingroup$
I dk why you mentioned that singletons are open in the discrete metric. An open ball $B_1/3(x_j)$ is, by definition, an open set containing $x_j,$ regardless of which metric.... I'm sure that in (a) it was also required that no $U_j $ is empty. And your work is OK in (a) although on a test or assignment I would include a proof, using the $triangle$ inequality, that $B_1/3(x_j)$ and $B_1/3(x_k)$ are disjoint when $jne k$ .
$endgroup$
– DanielWainfleet
Mar 14 at 9:47