Mixing
Prob. 1, Sec. 25, in Munkres' TOPOLOGY, 2nd ed: The components and the path components of $mathbbR$ with lower limit topology
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Here is Prob. 1, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
What are the components and the path components of $mathbbR_l$? What are the continuous maps $f colon mathbbR to mathbbR_l$?
My Attempt:
First of all, we note that $mathbbR_l$ denotes the set of real numbers with the topology having as a basis all the open intervals of the form $(a, b) = x in mathbbR colon a < x < b $, where $a$ and $b$ are any real numbers such that $a<b$, and all the closed-open intervals $[c, d) = c cup (c, d)$, where $c$ and $d$ are any real numbers such that $c < d$. Refer to Sec. 13 in Munkres.
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points. Let $a, b in A$ such that $a < b$. Then $a in (-infty, b)$ and $b in [b, +infty)$; these two rays are disjoint and their union is all of $mathbbR$; moreover these rays are open in $mathbbR_l$. So $A cap (-infty, b)$ and $A cap [b, +infty)$ is a separation of $A$. Thus $A$ is not connnected.
Thus we have shown that every set of real numbers having more than one point fails to be connected in $mathbbR_l$. In other words, the only nonempty connected sets in $mathbbR_l$ are the singleton sets of real numbers.
Since each component of $mathbbR_l$ must be connected [Refer to Theorem 25.1 in Munkres.], we can conclude from the preceding paragraph that each component of $mathbbR_l$ is a singleton set.
And, since each path component of $mathbbR_l$ must be contained in a component [Refer to Theorem 25.5 in Munkres.], we can conclude from the preceding paragraph that each path component of $mathbbR_l$ must be a singleton set too.
Finally let the map $f colon mathbbR to mathbbR_l$ be continuous. Then since $mathbbR$ is connected, so should be $f ( mathbbR)$ [Refer to Theorem 23.5 in Munkres.], and therefore $f ( mathbbR)$ must be a singleton set in $mathbbR_l$. This shows that $f$ is constant.
Is this solution correct? If so, is it clear enough too, especially for a beginning student? If not, then where are the issues?
general-topology proof-verification connectedness path-connected
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
add a comment |Â
up vote
1
down vote
favorite
Here is Prob. 1, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
What are the components and the path components of $mathbbR_l$? What are the continuous maps $f colon mathbbR to mathbbR_l$?
My Attempt:
First of all, we note that $mathbbR_l$ denotes the set of real numbers with the topology having as a basis all the open intervals of the form $(a, b) = x in mathbbR colon a < x < b $, where $a$ and $b$ are any real numbers such that $a<b$, and all the closed-open intervals $[c, d) = c cup (c, d)$, where $c$ and $d$ are any real numbers such that $c < d$. Refer to Sec. 13 in Munkres.
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points. Let $a, b in A$ such that $a < b$. Then $a in (-infty, b)$ and $b in [b, +infty)$; these two rays are disjoint and their union is all of $mathbbR$; moreover these rays are open in $mathbbR_l$. So $A cap (-infty, b)$ and $A cap [b, +infty)$ is a separation of $A$. Thus $A$ is not connnected.
Thus we have shown that every set of real numbers having more than one point fails to be connected in $mathbbR_l$. In other words, the only nonempty connected sets in $mathbbR_l$ are the singleton sets of real numbers.
Since each component of $mathbbR_l$ must be connected [Refer to Theorem 25.1 in Munkres.], we can conclude from the preceding paragraph that each component of $mathbbR_l$ is a singleton set.
And, since each path component of $mathbbR_l$ must be contained in a component [Refer to Theorem 25.5 in Munkres.], we can conclude from the preceding paragraph that each path component of $mathbbR_l$ must be a singleton set too.
Finally let the map $f colon mathbbR to mathbbR_l$ be continuous. Then since $mathbbR$ is connected, so should be $f ( mathbbR)$ [Refer to Theorem 23.5 in Munkres.], and therefore $f ( mathbbR)$ must be a singleton set in $mathbbR_l$. This shows that $f$ is constant.
Is this solution correct? If so, is it clear enough too, especially for a beginning student? If not, then where are the issues?
general-topology proof-verification connectedness path-connected
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is Prob. 1, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
What are the components and the path components of $mathbbR_l$? What are the continuous maps $f colon mathbbR to mathbbR_l$?
My Attempt:
First of all, we note that $mathbbR_l$ denotes the set of real numbers with the topology having as a basis all the open intervals of the form $(a, b) = x in mathbbR colon a < x < b $, where $a$ and $b$ are any real numbers such that $a<b$, and all the closed-open intervals $[c, d) = c cup (c, d)$, where $c$ and $d$ are any real numbers such that $c < d$. Refer to Sec. 13 in Munkres.
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points. Let $a, b in A$ such that $a < b$. Then $a in (-infty, b)$ and $b in [b, +infty)$; these two rays are disjoint and their union is all of $mathbbR$; moreover these rays are open in $mathbbR_l$. So $A cap (-infty, b)$ and $A cap [b, +infty)$ is a separation of $A$. Thus $A$ is not connnected.
Thus we have shown that every set of real numbers having more than one point fails to be connected in $mathbbR_l$. In other words, the only nonempty connected sets in $mathbbR_l$ are the singleton sets of real numbers.
Since each component of $mathbbR_l$ must be connected [Refer to Theorem 25.1 in Munkres.], we can conclude from the preceding paragraph that each component of $mathbbR_l$ is a singleton set.
And, since each path component of $mathbbR_l$ must be contained in a component [Refer to Theorem 25.5 in Munkres.], we can conclude from the preceding paragraph that each path component of $mathbbR_l$ must be a singleton set too.
Finally let the map $f colon mathbbR to mathbbR_l$ be continuous. Then since $mathbbR$ is connected, so should be $f ( mathbbR)$ [Refer to Theorem 23.5 in Munkres.], and therefore $f ( mathbbR)$ must be a singleton set in $mathbbR_l$. This shows that $f$ is constant.
Is this solution correct? If so, is it clear enough too, especially for a beginning student? If not, then where are the issues?
general-topology proof-verification connectedness path-connected
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
Here is Prob. 1, Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
What are the components and the path components of $mathbbR_l$? What are the continuous maps $f colon mathbbR to mathbbR_l$?
My Attempt:
First of all, we note that $mathbbR_l$ denotes the set of real numbers with the topology having as a basis all the open intervals of the form $(a, b) = x in mathbbR colon a < x < b $, where $a$ and $b$ are any real numbers such that $a<b$, and all the closed-open intervals $[c, d) = c cup (c, d)$, where $c$ and $d$ are any real numbers such that $c < d$. Refer to Sec. 13 in Munkres.
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points. Let $a, b in A$ such that $a < b$. Then $a in (-infty, b)$ and $b in [b, +infty)$; these two rays are disjoint and their union is all of $mathbbR$; moreover these rays are open in $mathbbR_l$. So $A cap (-infty, b)$ and $A cap [b, +infty)$ is a separation of $A$. Thus $A$ is not connnected.
Thus we have shown that every set of real numbers having more than one point fails to be connected in $mathbbR_l$. In other words, the only nonempty connected sets in $mathbbR_l$ are the singleton sets of real numbers.
Since each component of $mathbbR_l$ must be connected [Refer to Theorem 25.1 in Munkres.], we can conclude from the preceding paragraph that each component of $mathbbR_l$ is a singleton set.
And, since each path component of $mathbbR_l$ must be contained in a component [Refer to Theorem 25.5 in Munkres.], we can conclude from the preceding paragraph that each path component of $mathbbR_l$ must be a singleton set too.
Finally let the map $f colon mathbbR to mathbbR_l$ be continuous. Then since $mathbbR$ is connected, so should be $f ( mathbbR)$ [Refer to Theorem 23.5 in Munkres.], and therefore $f ( mathbbR)$ must be a singleton set in $mathbbR_l$. This shows that $f$ is constant.
Is this solution correct? If so, is it clear enough too, especially for a beginning student? If not, then where are the issues?
general-topology proof-verification connectedness path-connected
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
asked Jul 27 at 10:18
Saaqib Mahmood
7,11042169
7,11042169
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40
add a comment |Â
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Your solution is clear and accurate. Well-done!
This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points.
should instead read
. . . more than one point.
Apart from this, I have no comments for improvement.
answered Jul 27 at 10:26
Brahadeesh
3,40831246
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Your solution is clear and accurate. Well-done!
This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points.
should instead read
. . . more than one point.
Apart from this, I have no comments for improvement.
answered Jul 27 at 10:26
Brahadeesh
3,40831246
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
add a comment |Â
up vote
0
down vote
accepted
Your solution is clear and accurate. Well-done!
This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points.
should instead read
. . . more than one point.
Apart from this, I have no comments for improvement.
answered Jul 27 at 10:26
Brahadeesh
3,40831246
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Your solution is clear and accurate. Well-done!
This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points.
should instead read
. . . more than one point.
Apart from this, I have no comments for improvement.
answered Jul 27 at 10:26
Brahadeesh
3,40831246
Your solution is clear and accurate. Well-done!
This may just be a typo on your part, and if it is then I apologise for nitpicking, but the sentence
Now let $A$ be a set in $mathbbR_l$ consisting of more than one points.
should instead read
. . . more than one point.
Apart from this, I have no comments for improvement.
answered Jul 27 at 10:26
Brahadeesh
3,40831246
answered Jul 27 at 10:26
Brahadeesh
3,40831246
answered Jul 27 at 10:26
Brahadeesh
3,40831246
answered Jul 27 at 10:26
Brahadeesh
3,40831246
3,40831246
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
add a comment |Â
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
thank you for your answer. Can we connect using WhatsApp? My number is 0092 334 541 7958. I would like to talk to you about a lot of things in India!
– Saaqib Mahmood
Jul 27 at 11:00
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
@SaaqibMahmood sure, I’ve dropped you a message on WhatsApp. :)
– Brahadeesh
Jul 27 at 11:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864268%2fprob-1-sec-25-in-munkres-topology-2nd-ed-the-components-and-the-path-comp%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Exercise. Show if R has a base of only the closed-open intervals, then it has all the (a,b) intervals.
– William Elliot
Jul 27 at 10:39
@WilliamElliot we note that, for any two real numbers $a, b$ such that $a<b$, we have $$ (a, b) = bigcup_xin(a,b)[x,b). $$ Is this what you meant to point out?
– Saaqib Mahmood
Jul 27 at 11:03
Yes, that the definition is excessive.
– William Elliot
Jul 27 at 20:40