Show $(x,y) mapsto x+y$ is continuous from $Bbb R^n times Bbb R^n rightarrow Bbb R^n$
$begingroup$
If $mu$ is the map $(x,y) mapsto x+y$ from $Bbb R^n times Bbb R^n rightarrow Bbb R^n$, I'm trying to show that this is continuous(Assume product topology on $Bbb R^n times Bbb R^n$). I can see that I have to show that $mu^{-1}(V) in T_{Bbb R^n times Bbb R^n}$ if $V in T_{Bbb R^n}$. This means showing ultimately (since showing an element is in a union means showing it's in one set of the union) that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$, which means showing that $${(x,y) in Bbb R^n times Bbb R^n: |z_0-(x + y)| lt r_0 }=$$ $${(x,y) in Bbb R^n times Bbb R^n : |x_1-x|lt r_1 text{ and } |x_2-y| lt r_2} $$ for some $x_1, r_1, x_2, r_2$.
Showing $Leftarrow$ inclusion is simple by letting $x_1 + x_2 = z_0$ and letting $r_1=r_2 = frac{r_0}{2}$ and using the triangle inequality,
but I'm having trouble showing forward inclusion.
Anyone have any ideas?
real-analysis
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
$endgroup$
|
show 2 more comments
$begingroup$
If $mu$ is the map $(x,y) mapsto x+y$ from $Bbb R^n times Bbb R^n rightarrow Bbb R^n$, I'm trying to show that this is continuous(Assume product topology on $Bbb R^n times Bbb R^n$). I can see that I have to show that $mu^{-1}(V) in T_{Bbb R^n times Bbb R^n}$ if $V in T_{Bbb R^n}$. This means showing ultimately (since showing an element is in a union means showing it's in one set of the union) that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$, which means showing that $${(x,y) in Bbb R^n times Bbb R^n: |z_0-(x + y)| lt r_0 }=$$ $${(x,y) in Bbb R^n times Bbb R^n : |x_1-x|lt r_1 text{ and } |x_2-y| lt r_2} $$ for some $x_1, r_1, x_2, r_2$.
Showing $Leftarrow$ inclusion is simple by letting $x_1 + x_2 = z_0$ and letting $r_1=r_2 = frac{r_0}{2}$ and using the triangle inequality,
but I'm having trouble showing forward inclusion.
Anyone have any ideas?
real-analysis
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
$endgroup$
2
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25
|
show 2 more comments
$begingroup$
If $mu$ is the map $(x,y) mapsto x+y$ from $Bbb R^n times Bbb R^n rightarrow Bbb R^n$, I'm trying to show that this is continuous(Assume product topology on $Bbb R^n times Bbb R^n$). I can see that I have to show that $mu^{-1}(V) in T_{Bbb R^n times Bbb R^n}$ if $V in T_{Bbb R^n}$. This means showing ultimately (since showing an element is in a union means showing it's in one set of the union) that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$, which means showing that $${(x,y) in Bbb R^n times Bbb R^n: |z_0-(x + y)| lt r_0 }=$$ $${(x,y) in Bbb R^n times Bbb R^n : |x_1-x|lt r_1 text{ and } |x_2-y| lt r_2} $$ for some $x_1, r_1, x_2, r_2$.
Showing $Leftarrow$ inclusion is simple by letting $x_1 + x_2 = z_0$ and letting $r_1=r_2 = frac{r_0}{2}$ and using the triangle inequality,
but I'm having trouble showing forward inclusion.
Anyone have any ideas?
real-analysis
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
$endgroup$
If $mu$ is the map $(x,y) mapsto x+y$ from $Bbb R^n times Bbb R^n rightarrow Bbb R^n$, I'm trying to show that this is continuous(Assume product topology on $Bbb R^n times Bbb R^n$). I can see that I have to show that $mu^{-1}(V) in T_{Bbb R^n times Bbb R^n}$ if $V in T_{Bbb R^n}$. This means showing ultimately (since showing an element is in a union means showing it's in one set of the union) that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$, which means showing that $${(x,y) in Bbb R^n times Bbb R^n: |z_0-(x + y)| lt r_0 }=$$ $${(x,y) in Bbb R^n times Bbb R^n : |x_1-x|lt r_1 text{ and } |x_2-y| lt r_2} $$ for some $x_1, r_1, x_2, r_2$.
Showing $Leftarrow$ inclusion is simple by letting $x_1 + x_2 = z_0$ and letting $r_1=r_2 = frac{r_0}{2}$ and using the triangle inequality,
but I'm having trouble showing forward inclusion.
Anyone have any ideas?
real-analysis
real-analysis
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
edited Jan 20 at 8:11
Martin Sleziak
44.7k10118272
44.7k10118272
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
asked May 24 '17 at 12:22
Oliver GOliver G
1,4971531
1,4971531
2
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25
|
show 2 more comments
2
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25
2
2
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Define $f: mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}^n$ by $f(x,y) = x +y$. Let $B^n(f(x,y), epsilon)$ for some $epsilon > 0$ be a basic open set in $mathbb{R}^n$. It suffices to give a basic open set $(x,y) in B subset mathbb{R}^n times mathbb{R}^n$ such that $f(B) subset B^n (f(x,y),epsilon)$. Let $B=B^n(x,frac{epsilon}{2}) times B^n(y,frac{epsilon}{2})$, then for $(x_0, y_0) in B$ we have $|x-x_0|<frac{epsilon}{2}$ and $|y-y_0|< frac{epsilon}{2}$ thus $|f(x,y)-f(x_0,y_0)| = |x+y - x_0-y_0| leq |frac{epsilon}{2} + frac{epsilon}{2}|=epsilon$ and we have $f(B) subset B^n (f(x,y),epsilon)$ as desired.
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
$endgroup$
add a comment |
$begingroup$
Note that $f_1$ from $mathbb{R^n}×mathbb{R^n}$ to $mathbb{R^n}$ define by $f_1(x,y)=x$ for all $ x,y in $$mathbb{R^n}$ as $||f_1(x,y)-f_1(x',y')||leq ||(x,y)-(x',y')||$ hence continuous.
So similarly $f_2(x,y)=y$ for all $ x,y in $$mathbb{R^n}$ is continuous.
So $f(x,y)=f_1(x,y)+f_2(x,y)$ for all $ x,y in $$mathbb{R^n}$ is continuous.
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Define $f: mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}^n$ by $f(x,y) = x +y$. Let $B^n(f(x,y), epsilon)$ for some $epsilon > 0$ be a basic open set in $mathbb{R}^n$. It suffices to give a basic open set $(x,y) in B subset mathbb{R}^n times mathbb{R}^n$ such that $f(B) subset B^n (f(x,y),epsilon)$. Let $B=B^n(x,frac{epsilon}{2}) times B^n(y,frac{epsilon}{2})$, then for $(x_0, y_0) in B$ we have $|x-x_0|<frac{epsilon}{2}$ and $|y-y_0|< frac{epsilon}{2}$ thus $|f(x,y)-f(x_0,y_0)| = |x+y - x_0-y_0| leq |frac{epsilon}{2} + frac{epsilon}{2}|=epsilon$ and we have $f(B) subset B^n (f(x,y),epsilon)$ as desired.
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
$endgroup$
add a comment |
$begingroup$
Define $f: mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}^n$ by $f(x,y) = x +y$. Let $B^n(f(x,y), epsilon)$ for some $epsilon > 0$ be a basic open set in $mathbb{R}^n$. It suffices to give a basic open set $(x,y) in B subset mathbb{R}^n times mathbb{R}^n$ such that $f(B) subset B^n (f(x,y),epsilon)$. Let $B=B^n(x,frac{epsilon}{2}) times B^n(y,frac{epsilon}{2})$, then for $(x_0, y_0) in B$ we have $|x-x_0|<frac{epsilon}{2}$ and $|y-y_0|< frac{epsilon}{2}$ thus $|f(x,y)-f(x_0,y_0)| = |x+y - x_0-y_0| leq |frac{epsilon}{2} + frac{epsilon}{2}|=epsilon$ and we have $f(B) subset B^n (f(x,y),epsilon)$ as desired.
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
$endgroup$
add a comment |
$begingroup$
Define $f: mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}^n$ by $f(x,y) = x +y$. Let $B^n(f(x,y), epsilon)$ for some $epsilon > 0$ be a basic open set in $mathbb{R}^n$. It suffices to give a basic open set $(x,y) in B subset mathbb{R}^n times mathbb{R}^n$ such that $f(B) subset B^n (f(x,y),epsilon)$. Let $B=B^n(x,frac{epsilon}{2}) times B^n(y,frac{epsilon}{2})$, then for $(x_0, y_0) in B$ we have $|x-x_0|<frac{epsilon}{2}$ and $|y-y_0|< frac{epsilon}{2}$ thus $|f(x,y)-f(x_0,y_0)| = |x+y - x_0-y_0| leq |frac{epsilon}{2} + frac{epsilon}{2}|=epsilon$ and we have $f(B) subset B^n (f(x,y),epsilon)$ as desired.
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
$endgroup$
Define $f: mathbb{R}^n times mathbb{R}^n mapsto mathbb{R}^n$ by $f(x,y) = x +y$. Let $B^n(f(x,y), epsilon)$ for some $epsilon > 0$ be a basic open set in $mathbb{R}^n$. It suffices to give a basic open set $(x,y) in B subset mathbb{R}^n times mathbb{R}^n$ such that $f(B) subset B^n (f(x,y),epsilon)$. Let $B=B^n(x,frac{epsilon}{2}) times B^n(y,frac{epsilon}{2})$, then for $(x_0, y_0) in B$ we have $|x-x_0|<frac{epsilon}{2}$ and $|y-y_0|< frac{epsilon}{2}$ thus $|f(x,y)-f(x_0,y_0)| = |x+y - x_0-y_0| leq |frac{epsilon}{2} + frac{epsilon}{2}|=epsilon$ and we have $f(B) subset B^n (f(x,y),epsilon)$ as desired.
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
answered May 29 '17 at 1:26
Birch BryantBirch Bryant
1,622416
1,622416
add a comment |
add a comment |
$begingroup$
Note that $f_1$ from $mathbb{R^n}×mathbb{R^n}$ to $mathbb{R^n}$ define by $f_1(x,y)=x$ for all $ x,y in $$mathbb{R^n}$ as $||f_1(x,y)-f_1(x',y')||leq ||(x,y)-(x',y')||$ hence continuous.
So similarly $f_2(x,y)=y$ for all $ x,y in $$mathbb{R^n}$ is continuous.
So $f(x,y)=f_1(x,y)+f_2(x,y)$ for all $ x,y in $$mathbb{R^n}$ is continuous.
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
$endgroup$
add a comment |
$begingroup$
Note that $f_1$ from $mathbb{R^n}×mathbb{R^n}$ to $mathbb{R^n}$ define by $f_1(x,y)=x$ for all $ x,y in $$mathbb{R^n}$ as $||f_1(x,y)-f_1(x',y')||leq ||(x,y)-(x',y')||$ hence continuous.
So similarly $f_2(x,y)=y$ for all $ x,y in $$mathbb{R^n}$ is continuous.
So $f(x,y)=f_1(x,y)+f_2(x,y)$ for all $ x,y in $$mathbb{R^n}$ is continuous.
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
$endgroup$
add a comment |
$begingroup$
Note that $f_1$ from $mathbb{R^n}×mathbb{R^n}$ to $mathbb{R^n}$ define by $f_1(x,y)=x$ for all $ x,y in $$mathbb{R^n}$ as $||f_1(x,y)-f_1(x',y')||leq ||(x,y)-(x',y')||$ hence continuous.
So similarly $f_2(x,y)=y$ for all $ x,y in $$mathbb{R^n}$ is continuous.
So $f(x,y)=f_1(x,y)+f_2(x,y)$ for all $ x,y in $$mathbb{R^n}$ is continuous.
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
$endgroup$
Note that $f_1$ from $mathbb{R^n}×mathbb{R^n}$ to $mathbb{R^n}$ define by $f_1(x,y)=x$ for all $ x,y in $$mathbb{R^n}$ as $||f_1(x,y)-f_1(x',y')||leq ||(x,y)-(x',y')||$ hence continuous.
So similarly $f_2(x,y)=y$ for all $ x,y in $$mathbb{R^n}$ is continuous.
So $f(x,y)=f_1(x,y)+f_2(x,y)$ for all $ x,y in $$mathbb{R^n}$ is continuous.
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
edited Jan 20 at 9:50
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
answered Jan 20 at 7:58
Karambir.kdKarambir.kd
885
885
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2
$begingroup$
"This means showing ultimately that the inverse image of an open ball $B_{r_0}(x_0)$ can be written as a product of two open balls $B_{r_1}(x_1), B_{r_2}(x_2)$": it does not. It means showing that $mu^{-1}(V)$ can be written as the union of such products. For instance, in the case of $n = 1$, we'd like to be able to say that the open unit disk is an open set.
$endgroup$
– Omnomnomnom
May 24 '17 at 12:43
$begingroup$
@Omnomnomnom If $Bbb R^n times Bbb R^n$ has the product topology, then all open sets are of the form $U_1 times U_2$. So shouldn't $mu^{-1}(V) = U_1 times U_2$?
$endgroup$
– Oliver G
May 24 '17 at 12:56
$begingroup$
No. It is not the case that all open sets are of that form. Check the definition more closely.
$endgroup$
– Omnomnomnom
May 24 '17 at 13:00
$begingroup$
@Omnomnomnom Thank you for recognizing that, I went back and re-read the definition. So then, if $V = B_{r_0}(z_0)$, $A = {(x,y) in Bbb R^n times Bbb R^n : |z_0-(x+y)| lt r_0} = bigcup_i (B_{r_i}(a_i) times B_{p_i}(b_i)) = C$ is what I'm trying to prove. But to show an element $(x,y)$ of $A$ is also an element of $C$, I would have to show there exists a specific product of open balls in $C$ that contains $(x,y)$, and from here I'm stuck as in the question.
$endgroup$
– Oliver G
May 24 '17 at 15:02
$begingroup$
I'm assuming it's a triangle inequality relationship, but I can't seem to find one.
$endgroup$
– Oliver G
May 24 '17 at 15:25