Proving that given topological sets forms a partition.
$begingroup$
Let $A subset mathbb{R}$ be an open set.
Show that for a given $x in A$ the set
$$R_x = {y in A | (x,y] subset A}$$
is not-empty, open and path-connected. Conclude that $R_x$ is an open interval of the form $(x,r)$.
Show that the set
$$C_x = {y in A | [min{x,y},max{x,y}] subset A}$$
is an open interval and that for any $a,b in A$ holds $C_a = C_b$ or $C_a cap C_b = emptyset$.
By considering the rational numbers in ${C_x}_{x in A}$ conclude that one can write $$ A = cup_{i in mathbb{N}}(a_i,b_i) $$
where $a_i < b_i < a_{i+1}$.
This is what I got:
- Non-emptiness: Since $A$ is open we can just take an (suffieciently small) interval $[x,y]$ that lies completely in $A$. (The point $y ne x$ exists because $A$ is open and thus not a singleton.)
Openness: We observe that for all $y in R_x$ holds $(x,y] subseteq R_x$. Since $A$ is open we can find an open ball $B_epsilon (y) subset A$. Finally we observe $(x,y+epsilon /2] subset (x,y] cup B_epsilon (y) subset A$ and so $B_{epsilon /2}(y) subset R_x$.
Path connected: Since $(x,x] = emptyset subset A$ we see that for all $x,y in A$ holds $[x,y] subset A$. So the path connectedness is clear.
All in all we got that $R_x$ is open and connected and since it is in $mathbb{R}$ it is thus an open interval.
I have done $1)$ and $3)$ (under assumption that $2)$ holds) but I do not know how to prove $2)$. In particular I do not understand why it should hold that $C_a = C_b$ or $C_a cap C_b = emptyset$. Could you help me?
general-topology real-numbers
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
$endgroup$
add a comment |
$begingroup$
Let $A subset mathbb{R}$ be an open set.
Show that for a given $x in A$ the set
$$R_x = {y in A | (x,y] subset A}$$
is not-empty, open and path-connected. Conclude that $R_x$ is an open interval of the form $(x,r)$.
Show that the set
$$C_x = {y in A | [min{x,y},max{x,y}] subset A}$$
is an open interval and that for any $a,b in A$ holds $C_a = C_b$ or $C_a cap C_b = emptyset$.
By considering the rational numbers in ${C_x}_{x in A}$ conclude that one can write $$ A = cup_{i in mathbb{N}}(a_i,b_i) $$
where $a_i < b_i < a_{i+1}$.
This is what I got:
- Non-emptiness: Since $A$ is open we can just take an (suffieciently small) interval $[x,y]$ that lies completely in $A$. (The point $y ne x$ exists because $A$ is open and thus not a singleton.)
Openness: We observe that for all $y in R_x$ holds $(x,y] subseteq R_x$. Since $A$ is open we can find an open ball $B_epsilon (y) subset A$. Finally we observe $(x,y+epsilon /2] subset (x,y] cup B_epsilon (y) subset A$ and so $B_{epsilon /2}(y) subset R_x$.
Path connected: Since $(x,x] = emptyset subset A$ we see that for all $x,y in A$ holds $[x,y] subset A$. So the path connectedness is clear.
All in all we got that $R_x$ is open and connected and since it is in $mathbb{R}$ it is thus an open interval.
I have done $1)$ and $3)$ (under assumption that $2)$ holds) but I do not know how to prove $2)$. In particular I do not understand why it should hold that $C_a = C_b$ or $C_a cap C_b = emptyset$. Could you help me?
general-topology real-numbers
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
$endgroup$
$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54
add a comment |
$begingroup$
Let $A subset mathbb{R}$ be an open set.
Show that for a given $x in A$ the set
$$R_x = {y in A | (x,y] subset A}$$
is not-empty, open and path-connected. Conclude that $R_x$ is an open interval of the form $(x,r)$.
Show that the set
$$C_x = {y in A | [min{x,y},max{x,y}] subset A}$$
is an open interval and that for any $a,b in A$ holds $C_a = C_b$ or $C_a cap C_b = emptyset$.
By considering the rational numbers in ${C_x}_{x in A}$ conclude that one can write $$ A = cup_{i in mathbb{N}}(a_i,b_i) $$
where $a_i < b_i < a_{i+1}$.
This is what I got:
- Non-emptiness: Since $A$ is open we can just take an (suffieciently small) interval $[x,y]$ that lies completely in $A$. (The point $y ne x$ exists because $A$ is open and thus not a singleton.)
Openness: We observe that for all $y in R_x$ holds $(x,y] subseteq R_x$. Since $A$ is open we can find an open ball $B_epsilon (y) subset A$. Finally we observe $(x,y+epsilon /2] subset (x,y] cup B_epsilon (y) subset A$ and so $B_{epsilon /2}(y) subset R_x$.
Path connected: Since $(x,x] = emptyset subset A$ we see that for all $x,y in A$ holds $[x,y] subset A$. So the path connectedness is clear.
All in all we got that $R_x$ is open and connected and since it is in $mathbb{R}$ it is thus an open interval.
I have done $1)$ and $3)$ (under assumption that $2)$ holds) but I do not know how to prove $2)$. In particular I do not understand why it should hold that $C_a = C_b$ or $C_a cap C_b = emptyset$. Could you help me?
general-topology real-numbers
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
$endgroup$
Let $A subset mathbb{R}$ be an open set.
Show that for a given $x in A$ the set
$$R_x = {y in A | (x,y] subset A}$$
is not-empty, open and path-connected. Conclude that $R_x$ is an open interval of the form $(x,r)$.
Show that the set
$$C_x = {y in A | [min{x,y},max{x,y}] subset A}$$
is an open interval and that for any $a,b in A$ holds $C_a = C_b$ or $C_a cap C_b = emptyset$.
By considering the rational numbers in ${C_x}_{x in A}$ conclude that one can write $$ A = cup_{i in mathbb{N}}(a_i,b_i) $$
where $a_i < b_i < a_{i+1}$.
This is what I got:
- Non-emptiness: Since $A$ is open we can just take an (suffieciently small) interval $[x,y]$ that lies completely in $A$. (The point $y ne x$ exists because $A$ is open and thus not a singleton.)
Openness: We observe that for all $y in R_x$ holds $(x,y] subseteq R_x$. Since $A$ is open we can find an open ball $B_epsilon (y) subset A$. Finally we observe $(x,y+epsilon /2] subset (x,y] cup B_epsilon (y) subset A$ and so $B_{epsilon /2}(y) subset R_x$.
Path connected: Since $(x,x] = emptyset subset A$ we see that for all $x,y in A$ holds $[x,y] subset A$. So the path connectedness is clear.
All in all we got that $R_x$ is open and connected and since it is in $mathbb{R}$ it is thus an open interval.
I have done $1)$ and $3)$ (under assumption that $2)$ holds) but I do not know how to prove $2)$. In particular I do not understand why it should hold that $C_a = C_b$ or $C_a cap C_b = emptyset$. Could you help me?
general-topology real-numbers
general-topology real-numbers
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
edited Dec 6 '18 at 2:09
3nondatur
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
asked Dec 5 '18 at 22:58
3nondatur3nondatur
385111
385111
$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54
add a comment |
$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54
$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54
add a comment |
1 Answer
1
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$begingroup$
For 2.Obviously $ain C_asubset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $ain Jsubset A $ then $Jsubset C_a.$ (That is, $C_a$ is the $subseteq$-largest interval that contains $a$ and is a subset of $A.$)
Now if $P$ and $Q$ are intervals with non-empty intersection then $Pcup Q$ is an interval.
So suppose $C_a$ and $ C_b$ intersect. Let $J=C_acup C_b.$ Then $J$ is an interval, and $ain Jsubset A,$ so $Jsubset C_a .$ That is, $$C_acup C_b=Jsubset C_a$$ which implies $C_bsubset C_a.$
Interchanging $a,b$ in the preceding paragraph we also obtain $C_bsubset C_a.$ So $C_a=C_b$ if $C_acap C_bne emptyset.$
Remark: Let $In[a,y]=[min(a,y),max(a,y)]=[a,y]cup [y,a].$ If $J$ is an interval such that $ain Jsubset A$ then for every $yin J$ we have $yin A$ and $ In[a,y]subset Jsubset A,$ so $yin C_a.$ Hence $Jsubset C_a$.
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
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$begingroup$
For 2.Obviously $ain C_asubset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $ain Jsubset A $ then $Jsubset C_a.$ (That is, $C_a$ is the $subseteq$-largest interval that contains $a$ and is a subset of $A.$)
Now if $P$ and $Q$ are intervals with non-empty intersection then $Pcup Q$ is an interval.
So suppose $C_a$ and $ C_b$ intersect. Let $J=C_acup C_b.$ Then $J$ is an interval, and $ain Jsubset A,$ so $Jsubset C_a .$ That is, $$C_acup C_b=Jsubset C_a$$ which implies $C_bsubset C_a.$
Interchanging $a,b$ in the preceding paragraph we also obtain $C_bsubset C_a.$ So $C_a=C_b$ if $C_acap C_bne emptyset.$
Remark: Let $In[a,y]=[min(a,y),max(a,y)]=[a,y]cup [y,a].$ If $J$ is an interval such that $ain Jsubset A$ then for every $yin J$ we have $yin A$ and $ In[a,y]subset Jsubset A,$ so $yin C_a.$ Hence $Jsubset C_a$.
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
For 2.Obviously $ain C_asubset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $ain Jsubset A $ then $Jsubset C_a.$ (That is, $C_a$ is the $subseteq$-largest interval that contains $a$ and is a subset of $A.$)
Now if $P$ and $Q$ are intervals with non-empty intersection then $Pcup Q$ is an interval.
So suppose $C_a$ and $ C_b$ intersect. Let $J=C_acup C_b.$ Then $J$ is an interval, and $ain Jsubset A,$ so $Jsubset C_a .$ That is, $$C_acup C_b=Jsubset C_a$$ which implies $C_bsubset C_a.$
Interchanging $a,b$ in the preceding paragraph we also obtain $C_bsubset C_a.$ So $C_a=C_b$ if $C_acap C_bne emptyset.$
Remark: Let $In[a,y]=[min(a,y),max(a,y)]=[a,y]cup [y,a].$ If $J$ is an interval such that $ain Jsubset A$ then for every $yin J$ we have $yin A$ and $ In[a,y]subset Jsubset A,$ so $yin C_a.$ Hence $Jsubset C_a$.
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
add a comment |
$begingroup$
For 2.Obviously $ain C_asubset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $ain Jsubset A $ then $Jsubset C_a.$ (That is, $C_a$ is the $subseteq$-largest interval that contains $a$ and is a subset of $A.$)
Now if $P$ and $Q$ are intervals with non-empty intersection then $Pcup Q$ is an interval.
So suppose $C_a$ and $ C_b$ intersect. Let $J=C_acup C_b.$ Then $J$ is an interval, and $ain Jsubset A,$ so $Jsubset C_a .$ That is, $$C_acup C_b=Jsubset C_a$$ which implies $C_bsubset C_a.$
Interchanging $a,b$ in the preceding paragraph we also obtain $C_bsubset C_a.$ So $C_a=C_b$ if $C_acap C_bne emptyset.$
Remark: Let $In[a,y]=[min(a,y),max(a,y)]=[a,y]cup [y,a].$ If $J$ is an interval such that $ain Jsubset A$ then for every $yin J$ we have $yin A$ and $ In[a,y]subset Jsubset A,$ so $yin C_a.$ Hence $Jsubset C_a$.
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
$endgroup$
For 2.Obviously $ain C_asubset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $ain Jsubset A $ then $Jsubset C_a.$ (That is, $C_a$ is the $subseteq$-largest interval that contains $a$ and is a subset of $A.$)
Now if $P$ and $Q$ are intervals with non-empty intersection then $Pcup Q$ is an interval.
So suppose $C_a$ and $ C_b$ intersect. Let $J=C_acup C_b.$ Then $J$ is an interval, and $ain Jsubset A,$ so $Jsubset C_a .$ That is, $$C_acup C_b=Jsubset C_a$$ which implies $C_bsubset C_a.$
Interchanging $a,b$ in the preceding paragraph we also obtain $C_bsubset C_a.$ So $C_a=C_b$ if $C_acap C_bne emptyset.$
Remark: Let $In[a,y]=[min(a,y),max(a,y)]=[a,y]cup [y,a].$ If $J$ is an interval such that $ain Jsubset A$ then for every $yin J$ we have $yin A$ and $ In[a,y]subset Jsubset A,$ so $yin C_a.$ Hence $Jsubset C_a$.
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
edited Dec 6 '18 at 8:53
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
answered Dec 6 '18 at 8:30
DanielWainfleetDanielWainfleet
34.6k31648
34.6k31648
add a comment |
add a comment |
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$begingroup$
You are obviously misquoting 2). Any two sets are either equal or unequal!
$endgroup$
– Kavi Rama Murthy
Dec 5 '18 at 23:38
$begingroup$
You are right, I corrected it.
$endgroup$
– 3nondatur
Dec 5 '18 at 23:53
$begingroup$
[x,y] or [y,x] cannot be in A. A is a bunch of points and neither of them is a point. They are subsets of A.
$endgroup$
– William Elliot
Dec 6 '18 at 2:15
$begingroup$
@WilliamElliot Yes, silly me. It's of course $[min(x,y),max(x,y)]subset A$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 7:22
$begingroup$
Note that in this context "interval" means "convex set". E.g. $Bbb R$ and $emptyset$ and $(0,infty)$ are intervals
$endgroup$
– DanielWainfleet
Dec 6 '18 at 8:54