Help with completing proof of cartesian products of two path connected spaces being path connected.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












I was wondering if somebody can help me with completing a proof of the claim made above. The problem I am having occurs at the end of the proof, so I don't want to bring it up over here without context. I would like to add that this is not homework for any class. I'm studying topology on my own for fun.



I apologise if you find that proving this claim instead of for arbitary cartesian products of path connected spaces is particularly reductive. I intend to prove that next. I just wanted to prove this as a lemma first in order to get a feel for the proof of the larger claim. Also, please let me know if you need to know any definitions.



With all of that out of the way, here is what I have thus far:



Recall: A space X is said to be path connected if every pair of points of X can be joined by a path in X.



Let X and Y be path connected topological spaces. Suppose they are non-empty, for the result is trivial otherwise. Let $x_1$ and $x_2$ be elements of X and let $y_1$ and $y_2$ be elements of Y. We are interested in showing that there is a path from $x_1$ x $y_1$ to $x_2$ x $y_2$. There exists a continuous map $f: [a,b] rightarrow X$ of some closed interval of $mathbbR$ into X, such that $f(a) = x_1$ and $f(b) = x_2$ because X is path connected. There also exists a continuous map $g: [c,d] rightarrow Y$ of some closed interval of $mathbbR$ into Y, such that $f(c) = y_1$ and $f(d) = y_2$ because Y is path connected.



Let $A = [a,b]$ x $[c,d]$. Define $f': A rightarrow X$ by $f'(x) = f(pi_1(x)) = f circ pi_1 (x)$ and $g': A rightarrow Y$ by $g'(x) = g(pi_2(x)) = g circ pi_2 (x)$.



Are these projections onto the first and second factors ($pi_1$ and $pi_2$, respectively) continuous? We'll start with $pi_1: $ X x Y $ rightarrow X$, where X and Y are arbitrary topological spaces. Let U be open in X. We want to show that $pi_1 ^-1(U)$ is open in X x Y. Notice that $pi _ 1$ need not be injective, but it is certainly surjective. Notice that U x Y $subset pi _ 1^-1(U)$ is open in X x Y. Is it true that $pi _1^-1(U) subset$ U x Y? Fix u x v $in pi _ 1^-1(U)$. It is clear that $u in U$, for if it wasn't, then $pi _1 ($u x v$) notin U$ (which cannot be). That $v in Y$ is trivial. Hence, u x v $in$ U x Y and $pi _ 1^-1(U) subset$ U x Y. Hence, $pi _ 1^-1(U) = $ U x Y. We conclude that $pi _ 1$ is continuous. Similarly, $pi _2$ is continuous.



Knowing that, since $f$ and $g$ are continuous, we note that $f'$ and $g'$ are continuous by Theorem 18.2(c).



Let $h: A rightarrow$ X x Y be given by the equation $h(a) = (f'(a),g'(a))$. We conclude that $h$ is continuous by Theorem 18.4 (Maps into products).



The problem I am having is whether or not [a,b] x [c,d] = [a x c, b x d]. This is needed to show that h is a path in X x Y.



This might be a particularly stupid thing to get caught up on, but I feel like I'm missing something here. I don't know what the order relation for the latter set is (I am aware, of course, that it is a subset of $mathbbR$ x $mathbbR$). Is it the dictionary order relation?



If anybody has any comments on the proof (mistakes, improvements, etc), please let me know. I'm fairly new to higher-level math (I'm a physics student) as I have only had a year of exposure to rigorous math, so any criticism is greatly appreciated. :) Also, thanks in advance for the help!



Theorems:



Theorem 18.2(c): Let X,Y, and Z be topological spaces. If $f: X rightarrow Y$ and $g: Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.



Theorem 18.4 (Maps into products): Let $f: A rightarrow$ X x Y be given by the equation $$f(a) = (f_1(a),f_2(1)).$$ Then $f$ is continuous iff the functions $f_1: A rightarrow X$ and $f_2: A rightarrow Y$ are continuous.




general-topology path-connected




share|cite|improve this question















  • 1




    What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
    – bof
    Aug 1 at 0:27











  • Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
    – TenaxPropositi
    Aug 1 at 0:33










  • would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
    – TenaxPropositi
    Aug 1 at 0:33











  • Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
    – bof
    Aug 1 at 0:44











  • OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
    – bof
    Aug 1 at 0:48















up vote
1
down vote

favorite












I was wondering if somebody can help me with completing a proof of the claim made above. The problem I am having occurs at the end of the proof, so I don't want to bring it up over here without context. I would like to add that this is not homework for any class. I'm studying topology on my own for fun.



I apologise if you find that proving this claim instead of for arbitary cartesian products of path connected spaces is particularly reductive. I intend to prove that next. I just wanted to prove this as a lemma first in order to get a feel for the proof of the larger claim. Also, please let me know if you need to know any definitions.



With all of that out of the way, here is what I have thus far:



Recall: A space X is said to be path connected if every pair of points of X can be joined by a path in X.



Let X and Y be path connected topological spaces. Suppose they are non-empty, for the result is trivial otherwise. Let $x_1$ and $x_2$ be elements of X and let $y_1$ and $y_2$ be elements of Y. We are interested in showing that there is a path from $x_1$ x $y_1$ to $x_2$ x $y_2$. There exists a continuous map $f: [a,b] rightarrow X$ of some closed interval of $mathbbR$ into X, such that $f(a) = x_1$ and $f(b) = x_2$ because X is path connected. There also exists a continuous map $g: [c,d] rightarrow Y$ of some closed interval of $mathbbR$ into Y, such that $f(c) = y_1$ and $f(d) = y_2$ because Y is path connected.



Let $A = [a,b]$ x $[c,d]$. Define $f': A rightarrow X$ by $f'(x) = f(pi_1(x)) = f circ pi_1 (x)$ and $g': A rightarrow Y$ by $g'(x) = g(pi_2(x)) = g circ pi_2 (x)$.



Are these projections onto the first and second factors ($pi_1$ and $pi_2$, respectively) continuous? We'll start with $pi_1: $ X x Y $ rightarrow X$, where X and Y are arbitrary topological spaces. Let U be open in X. We want to show that $pi_1 ^-1(U)$ is open in X x Y. Notice that $pi _ 1$ need not be injective, but it is certainly surjective. Notice that U x Y $subset pi _ 1^-1(U)$ is open in X x Y. Is it true that $pi _1^-1(U) subset$ U x Y? Fix u x v $in pi _ 1^-1(U)$. It is clear that $u in U$, for if it wasn't, then $pi _1 ($u x v$) notin U$ (which cannot be). That $v in Y$ is trivial. Hence, u x v $in$ U x Y and $pi _ 1^-1(U) subset$ U x Y. Hence, $pi _ 1^-1(U) = $ U x Y. We conclude that $pi _ 1$ is continuous. Similarly, $pi _2$ is continuous.



Knowing that, since $f$ and $g$ are continuous, we note that $f'$ and $g'$ are continuous by Theorem 18.2(c).



Let $h: A rightarrow$ X x Y be given by the equation $h(a) = (f'(a),g'(a))$. We conclude that $h$ is continuous by Theorem 18.4 (Maps into products).



The problem I am having is whether or not [a,b] x [c,d] = [a x c, b x d]. This is needed to show that h is a path in X x Y.



This might be a particularly stupid thing to get caught up on, but I feel like I'm missing something here. I don't know what the order relation for the latter set is (I am aware, of course, that it is a subset of $mathbbR$ x $mathbbR$). Is it the dictionary order relation?



If anybody has any comments on the proof (mistakes, improvements, etc), please let me know. I'm fairly new to higher-level math (I'm a physics student) as I have only had a year of exposure to rigorous math, so any criticism is greatly appreciated. :) Also, thanks in advance for the help!



Theorems:



Theorem 18.2(c): Let X,Y, and Z be topological spaces. If $f: X rightarrow Y$ and $g: Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.



Theorem 18.4 (Maps into products): Let $f: A rightarrow$ X x Y be given by the equation $$f(a) = (f_1(a),f_2(1)).$$ Then $f$ is continuous iff the functions $f_1: A rightarrow X$ and $f_2: A rightarrow Y$ are continuous.




general-topology path-connected




share|cite|improve this question















  • 1




    What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
    – bof
    Aug 1 at 0:27











  • Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
    – TenaxPropositi
    Aug 1 at 0:33










  • would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
    – TenaxPropositi
    Aug 1 at 0:33











  • Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
    – bof
    Aug 1 at 0:44











  • OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
    – bof
    Aug 1 at 0:48













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was wondering if somebody can help me with completing a proof of the claim made above. The problem I am having occurs at the end of the proof, so I don't want to bring it up over here without context. I would like to add that this is not homework for any class. I'm studying topology on my own for fun.



I apologise if you find that proving this claim instead of for arbitary cartesian products of path connected spaces is particularly reductive. I intend to prove that next. I just wanted to prove this as a lemma first in order to get a feel for the proof of the larger claim. Also, please let me know if you need to know any definitions.



With all of that out of the way, here is what I have thus far:



Recall: A space X is said to be path connected if every pair of points of X can be joined by a path in X.



Let X and Y be path connected topological spaces. Suppose they are non-empty, for the result is trivial otherwise. Let $x_1$ and $x_2$ be elements of X and let $y_1$ and $y_2$ be elements of Y. We are interested in showing that there is a path from $x_1$ x $y_1$ to $x_2$ x $y_2$. There exists a continuous map $f: [a,b] rightarrow X$ of some closed interval of $mathbbR$ into X, such that $f(a) = x_1$ and $f(b) = x_2$ because X is path connected. There also exists a continuous map $g: [c,d] rightarrow Y$ of some closed interval of $mathbbR$ into Y, such that $f(c) = y_1$ and $f(d) = y_2$ because Y is path connected.



Let $A = [a,b]$ x $[c,d]$. Define $f': A rightarrow X$ by $f'(x) = f(pi_1(x)) = f circ pi_1 (x)$ and $g': A rightarrow Y$ by $g'(x) = g(pi_2(x)) = g circ pi_2 (x)$.



Are these projections onto the first and second factors ($pi_1$ and $pi_2$, respectively) continuous? We'll start with $pi_1: $ X x Y $ rightarrow X$, where X and Y are arbitrary topological spaces. Let U be open in X. We want to show that $pi_1 ^-1(U)$ is open in X x Y. Notice that $pi _ 1$ need not be injective, but it is certainly surjective. Notice that U x Y $subset pi _ 1^-1(U)$ is open in X x Y. Is it true that $pi _1^-1(U) subset$ U x Y? Fix u x v $in pi _ 1^-1(U)$. It is clear that $u in U$, for if it wasn't, then $pi _1 ($u x v$) notin U$ (which cannot be). That $v in Y$ is trivial. Hence, u x v $in$ U x Y and $pi _ 1^-1(U) subset$ U x Y. Hence, $pi _ 1^-1(U) = $ U x Y. We conclude that $pi _ 1$ is continuous. Similarly, $pi _2$ is continuous.



Knowing that, since $f$ and $g$ are continuous, we note that $f'$ and $g'$ are continuous by Theorem 18.2(c).



Let $h: A rightarrow$ X x Y be given by the equation $h(a) = (f'(a),g'(a))$. We conclude that $h$ is continuous by Theorem 18.4 (Maps into products).



The problem I am having is whether or not [a,b] x [c,d] = [a x c, b x d]. This is needed to show that h is a path in X x Y.



This might be a particularly stupid thing to get caught up on, but I feel like I'm missing something here. I don't know what the order relation for the latter set is (I am aware, of course, that it is a subset of $mathbbR$ x $mathbbR$). Is it the dictionary order relation?



If anybody has any comments on the proof (mistakes, improvements, etc), please let me know. I'm fairly new to higher-level math (I'm a physics student) as I have only had a year of exposure to rigorous math, so any criticism is greatly appreciated. :) Also, thanks in advance for the help!



Theorems:



Theorem 18.2(c): Let X,Y, and Z be topological spaces. If $f: X rightarrow Y$ and $g: Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.



Theorem 18.4 (Maps into products): Let $f: A rightarrow$ X x Y be given by the equation $$f(a) = (f_1(a),f_2(1)).$$ Then $f$ is continuous iff the functions $f_1: A rightarrow X$ and $f_2: A rightarrow Y$ are continuous.




general-topology path-connected




share|cite|improve this question











I was wondering if somebody can help me with completing a proof of the claim made above. The problem I am having occurs at the end of the proof, so I don't want to bring it up over here without context. I would like to add that this is not homework for any class. I'm studying topology on my own for fun.



I apologise if you find that proving this claim instead of for arbitary cartesian products of path connected spaces is particularly reductive. I intend to prove that next. I just wanted to prove this as a lemma first in order to get a feel for the proof of the larger claim. Also, please let me know if you need to know any definitions.



With all of that out of the way, here is what I have thus far:



Recall: A space X is said to be path connected if every pair of points of X can be joined by a path in X.



Let X and Y be path connected topological spaces. Suppose they are non-empty, for the result is trivial otherwise. Let $x_1$ and $x_2$ be elements of X and let $y_1$ and $y_2$ be elements of Y. We are interested in showing that there is a path from $x_1$ x $y_1$ to $x_2$ x $y_2$. There exists a continuous map $f: [a,b] rightarrow X$ of some closed interval of $mathbbR$ into X, such that $f(a) = x_1$ and $f(b) = x_2$ because X is path connected. There also exists a continuous map $g: [c,d] rightarrow Y$ of some closed interval of $mathbbR$ into Y, such that $f(c) = y_1$ and $f(d) = y_2$ because Y is path connected.



Let $A = [a,b]$ x $[c,d]$. Define $f': A rightarrow X$ by $f'(x) = f(pi_1(x)) = f circ pi_1 (x)$ and $g': A rightarrow Y$ by $g'(x) = g(pi_2(x)) = g circ pi_2 (x)$.



Are these projections onto the first and second factors ($pi_1$ and $pi_2$, respectively) continuous? We'll start with $pi_1: $ X x Y $ rightarrow X$, where X and Y are arbitrary topological spaces. Let U be open in X. We want to show that $pi_1 ^-1(U)$ is open in X x Y. Notice that $pi _ 1$ need not be injective, but it is certainly surjective. Notice that U x Y $subset pi _ 1^-1(U)$ is open in X x Y. Is it true that $pi _1^-1(U) subset$ U x Y? Fix u x v $in pi _ 1^-1(U)$. It is clear that $u in U$, for if it wasn't, then $pi _1 ($u x v$) notin U$ (which cannot be). That $v in Y$ is trivial. Hence, u x v $in$ U x Y and $pi _ 1^-1(U) subset$ U x Y. Hence, $pi _ 1^-1(U) = $ U x Y. We conclude that $pi _ 1$ is continuous. Similarly, $pi _2$ is continuous.



Knowing that, since $f$ and $g$ are continuous, we note that $f'$ and $g'$ are continuous by Theorem 18.2(c).



Let $h: A rightarrow$ X x Y be given by the equation $h(a) = (f'(a),g'(a))$. We conclude that $h$ is continuous by Theorem 18.4 (Maps into products).



The problem I am having is whether or not [a,b] x [c,d] = [a x c, b x d]. This is needed to show that h is a path in X x Y.



This might be a particularly stupid thing to get caught up on, but I feel like I'm missing something here. I don't know what the order relation for the latter set is (I am aware, of course, that it is a subset of $mathbbR$ x $mathbbR$). Is it the dictionary order relation?



If anybody has any comments on the proof (mistakes, improvements, etc), please let me know. I'm fairly new to higher-level math (I'm a physics student) as I have only had a year of exposure to rigorous math, so any criticism is greatly appreciated. :) Also, thanks in advance for the help!



Theorems:



Theorem 18.2(c): Let X,Y, and Z be topological spaces. If $f: X rightarrow Y$ and $g: Y rightarrow Z$ are continuous, then the map $g circ f: X rightarrow Z$ is continuous.



Theorem 18.4 (Maps into products): Let $f: A rightarrow$ X x Y be given by the equation $$f(a) = (f_1(a),f_2(1)).$$ Then $f$ is continuous iff the functions $f_1: A rightarrow X$ and $f_2: A rightarrow Y$ are continuous.





general-topology path-connected





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 1 at 0:15









TenaxPropositi

84




84







  • 1




    What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
    – bof
    Aug 1 at 0:27











  • Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
    – TenaxPropositi
    Aug 1 at 0:33










  • would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
    – TenaxPropositi
    Aug 1 at 0:33











  • Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
    – bof
    Aug 1 at 0:44











  • OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
    – bof
    Aug 1 at 0:48













  • 1




    What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
    – bof
    Aug 1 at 0:27











  • Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
    – TenaxPropositi
    Aug 1 at 0:33










  • would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
    – TenaxPropositi
    Aug 1 at 0:33











  • Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
    – bof
    Aug 1 at 0:44











  • OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
    – bof
    Aug 1 at 0:48








1




1




What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
– bof
Aug 1 at 0:27





What's the point of confusing the issue with $a,b,c,d$? Why don't you just take $f:[0,1]to X$ and $g:[0,1]to Y$? And why would you want to define a function on $[a,b]times[c,d]$? Why don't you just define $h:[0,1]to Xtimes Y$ by $h(t)=(f(t),g(t))$? Also, why use the weird notation $xtimes y$ for an element of $Xtimes Y$ instead of the normal $(x,y)$?
– bof
Aug 1 at 0:27













Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
– TenaxPropositi
Aug 1 at 0:33




Hi bof! Why would I take f and g to be those two, though? X and Y are arbitrary path connected spaces. Perhaps if I provide the definition of a path in my book, things might be clearer. A path in a space X from x to y is a continuous map f: [a,b] --> X of some closed interval in the real line into X, such that f(a) = x and f(b) = y. I wanted these x and y to be entirely arbitrary, and consequently I can't choose what a and b are. Perhaps I am missing something though. I don't want to define a function on that in and of itself. I chose A to a product of closed intervals and was hoping it
– TenaxPropositi
Aug 1 at 0:33












would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
– TenaxPropositi
Aug 1 at 0:33





would equal a closed interval. I'm sorry about the notation being wierd. I don't know what all of the standard practices in topology are. That is merely what was used in my book (Munkres).
– TenaxPropositi
Aug 1 at 0:33













Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
– bof
Aug 1 at 0:44





Actually, you can choose what $a$ and $b$ are. That';s why most books define a path from $x_1$ to $x_2$ as a continuous function $f:[0,1]to X$ such that $f(0)=x_1$ and $f(1)=x_2.$ Anyway, if $f:[a,b]to X$ is a path from $x_1$ to $x_2,$ then so is $g:[0,1]to X$ where $g(t)=f(a+t(b-a)).$
– bof
Aug 1 at 0:44













OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
– bof
Aug 1 at 0:48





OK, how about this. Given paths $f:[a,b]to X$ and $g:[c,d]to Y,$ define a path $h:[0,1]to Xtimes Y$ by setting $h(t)=f(a+t(b-a))times g(c+t(d-c)).$
– bof
Aug 1 at 0:48











1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Let $(x,y)$ and $(z,w)$ be two points in $Xtimes Y.$ Since $X$ is path-connected $Xtimes y$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$.
By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $ztimes Y$ is path connected. Concatenating $f$ and $g$ you get the required path.






share|cite|improve this answer





















  • Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
    – TenaxPropositi
    2 days ago











  • Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
    – TenaxPropositi
    2 days ago










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868600%2fhelp-with-completing-proof-of-cartesian-products-of-two-path-connected-spaces-be%23new-answer', 'question_page');

);

Post as a guest

















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 $(x,y)$ and $(z,w)$ be two points in $Xtimes Y.$ Since $X$ is path-connected $Xtimes y$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$.
By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $ztimes Y$ is path connected. Concatenating $f$ and $g$ you get the required path.






share|cite|improve this answer





















  • Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
    – TenaxPropositi
    2 days ago











  • Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
    – TenaxPropositi
    2 days ago














up vote
1
down vote



accepted










Let $(x,y)$ and $(z,w)$ be two points in $Xtimes Y.$ Since $X$ is path-connected $Xtimes y$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$.
By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $ztimes Y$ is path connected. Concatenating $f$ and $g$ you get the required path.






share|cite|improve this answer





















  • Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
    – TenaxPropositi
    2 days ago











  • Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
    – TenaxPropositi
    2 days ago












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Let $(x,y)$ and $(z,w)$ be two points in $Xtimes Y.$ Since $X$ is path-connected $Xtimes y$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$.
By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $ztimes Y$ is path connected. Concatenating $f$ and $g$ you get the required path.






share|cite|improve this answer













Let $(x,y)$ and $(z,w)$ be two points in $Xtimes Y.$ Since $X$ is path-connected $Xtimes y$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$.
By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $ztimes Y$ is path connected. Concatenating $f$ and $g$ you get the required path.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 4 at 21:05









user345777

23529




23529











  • Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
    – TenaxPropositi
    2 days ago











  • Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
    – TenaxPropositi
    2 days ago
















  • Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
    – TenaxPropositi
    2 days ago











  • Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
    – TenaxPropositi
    2 days ago















Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
– TenaxPropositi
2 days ago





Hi! Thanks your response! I'm not sure whether or not your question was rhetorical but here's my attempt: Fix (x,y), (z,y) in X x y. There exists a path in X f: [a,b] --> X s.t. f(a) = x and f(b) = y by the path connectedness of X. Let g:[a,b] --> y be defined by g(t) = y. g is continuous by theorem 18.2 of Munkres (Let X,Y be topological spaces. If f: X --> Y maps all of X into the single point y0 of Y, then f is continuous). Then h:[a,b] --> X x y defined by h(t) = f(t) x g(t) is continuous by thm 18.4 (see theorems at the end of the question) and is a path.
– TenaxPropositi
2 days ago













Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
– TenaxPropositi
2 days ago




Sorry, I meant f:[a,b] --> X s.t. f(a) = x and f(b) = z. My apologies.
– TenaxPropositi
2 days ago












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868600%2fhelp-with-completing-proof-of-cartesian-products-of-two-path-connected-spaces-be%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more