Continuous maps between $mathbb Ncup {infty}$ and $mathbb R$
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Consider the set $N=mathbb Ncup{infty}$ together with the following topology: a subset $U$ of $N$ is open if either $inftynotin U$ or $Nsetminus U$ is finite.
(1) Describe continuous maps $mathbb Rto N$ and $Nto mathbb R$.
(2) Does there exist a subset of $mathbb R$ homeomorphic to $N$?
--
(1) I'm not quite sure what is being asked. A continuous map is one with the property that preimages of open sets are open. We know how open sets look like in both spaces. But what exactly can I conclude about continuous maps?
(2) It looks like $N$ is compact. So the only candidates for such subsets are compact subsets of $mathbb R$. But I guess I need to understand (1) first? If I do, I will have understand how restrictions of continuous maps look like as well, I suppose.
general-topology
asked Dec 3 '18 at 1:40
user531587user531587
261113
$endgroup$
add a comment |
$begingroup$
Consider the set $N=mathbb Ncup{infty}$ together with the following topology: a subset $U$ of $N$ is open if either $inftynotin U$ or $Nsetminus U$ is finite.
(1) Describe continuous maps $mathbb Rto N$ and $Nto mathbb R$.
(2) Does there exist a subset of $mathbb R$ homeomorphic to $N$?
--
(1) I'm not quite sure what is being asked. A continuous map is one with the property that preimages of open sets are open. We know how open sets look like in both spaces. But what exactly can I conclude about continuous maps?
(2) It looks like $N$ is compact. So the only candidates for such subsets are compact subsets of $mathbb R$. But I guess I need to understand (1) first? If I do, I will have understand how restrictions of continuous maps look like as well, I suppose.
general-topology
asked Dec 3 '18 at 1:40
user531587user531587
261113
$endgroup$
$begingroup$
Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
$endgroup$
– Luca Carai
Dec 3 '18 at 1:57
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@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
$endgroup$
– user531587
Dec 3 '18 at 2:19
$begingroup$
Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
$endgroup$
– Joel Pereira
Dec 3 '18 at 2:28
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@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
$endgroup$
– user531587
Dec 3 '18 at 3:08
add a comment |
$begingroup$
Consider the set $N=mathbb Ncup{infty}$ together with the following topology: a subset $U$ of $N$ is open if either $inftynotin U$ or $Nsetminus U$ is finite.
(1) Describe continuous maps $mathbb Rto N$ and $Nto mathbb R$.
(2) Does there exist a subset of $mathbb R$ homeomorphic to $N$?
--
(1) I'm not quite sure what is being asked. A continuous map is one with the property that preimages of open sets are open. We know how open sets look like in both spaces. But what exactly can I conclude about continuous maps?
(2) It looks like $N$ is compact. So the only candidates for such subsets are compact subsets of $mathbb R$. But I guess I need to understand (1) first? If I do, I will have understand how restrictions of continuous maps look like as well, I suppose.
general-topology
asked Dec 3 '18 at 1:40
user531587user531587
261113
$endgroup$
Consider the set $N=mathbb Ncup{infty}$ together with the following topology: a subset $U$ of $N$ is open if either $inftynotin U$ or $Nsetminus U$ is finite.
(1) Describe continuous maps $mathbb Rto N$ and $Nto mathbb R$.
(2) Does there exist a subset of $mathbb R$ homeomorphic to $N$?
--
(1) I'm not quite sure what is being asked. A continuous map is one with the property that preimages of open sets are open. We know how open sets look like in both spaces. But what exactly can I conclude about continuous maps?
(2) It looks like $N$ is compact. So the only candidates for such subsets are compact subsets of $mathbb R$. But I guess I need to understand (1) first? If I do, I will have understand how restrictions of continuous maps look like as well, I suppose.
general-topology
general-topology
asked Dec 3 '18 at 1:40
user531587user531587
261113
asked Dec 3 '18 at 1:40
user531587user531587
261113
asked Dec 3 '18 at 1:40
user531587user531587
261113
asked Dec 3 '18 at 1:40
user531587user531587
261113
asked Dec 3 '18 at 1:40
user531587user531587
261113
261113
$begingroup$
Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
$endgroup$
– Luca Carai
Dec 3 '18 at 1:57
$begingroup$
@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
$endgroup$
– user531587
Dec 3 '18 at 2:19
$begingroup$
Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
$endgroup$
– Joel Pereira
Dec 3 '18 at 2:28
$begingroup$
@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
$endgroup$
– user531587
Dec 3 '18 at 3:08
add a comment |
$begingroup$
Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
$endgroup$
– Luca Carai
Dec 3 '18 at 1:57
$begingroup$
@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
$endgroup$
– user531587
Dec 3 '18 at 2:19
$begingroup$
Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
$endgroup$
– Joel Pereira
Dec 3 '18 at 2:28
$begingroup$
@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
$endgroup$
– user531587
Dec 3 '18 at 3:08
$begingroup$
Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
$endgroup$
– Luca Carai
Dec 3 '18 at 1:57
$begingroup$
Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
$endgroup$
– Luca Carai
Dec 3 '18 at 1:57
$begingroup$
@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
$endgroup$
– user531587
Dec 3 '18 at 2:19
$begingroup$
@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
$endgroup$
– user531587
Dec 3 '18 at 2:19
$begingroup$
Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
$endgroup$
– Joel Pereira
Dec 3 '18 at 2:28
$begingroup$
Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
$endgroup$
– Joel Pereira
Dec 3 '18 at 2:28
$begingroup$
@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
$endgroup$
– user531587
Dec 3 '18 at 3:08
$begingroup$
@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
$endgroup$
– user531587
Dec 3 '18 at 3:08
add a comment |
1 Answer
1
active
oldest
votes
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(1) a continuous map $f$ from $mathbb{N} cup {infty}$ into $mathbb{R}$ corresponds to a convergent sequence and its limit, in the sense that for any space $X$, $f: mathbb{N} cup {infty} to X$ is continuous iff $x_n = f(n)$ defines a sequence that converges to $f(infty)$ in $X$. And conversely for every sequence $x_n$ in $X$ that converges to $x$, the function defined by $f(n) = x_n$ for all $n$ and $f(infty) = x$, is continuous from $mathbb{N} cup {infty}$ to $X$.
The other way around (from $mathbb{R}$ to $mathbb{N} cup {infty}$ there are only constant maps as $mathbb{N} cup {infty}$ is totally disconnected, and $mathbb{R}$ is connected and thus has connected image.
Any convergent sequence with limit (like ${frac{1}{n}: n ge 1} cup {0}$) is homeomorphic to $mathbb{N} cup {infty}$, as is easily checked.
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
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For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
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– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
$endgroup$
– user531587
Dec 5 '18 at 3:55
$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57
add a comment |
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$begingroup$
(1) a continuous map $f$ from $mathbb{N} cup {infty}$ into $mathbb{R}$ corresponds to a convergent sequence and its limit, in the sense that for any space $X$, $f: mathbb{N} cup {infty} to X$ is continuous iff $x_n = f(n)$ defines a sequence that converges to $f(infty)$ in $X$. And conversely for every sequence $x_n$ in $X$ that converges to $x$, the function defined by $f(n) = x_n$ for all $n$ and $f(infty) = x$, is continuous from $mathbb{N} cup {infty}$ to $X$.
The other way around (from $mathbb{R}$ to $mathbb{N} cup {infty}$ there are only constant maps as $mathbb{N} cup {infty}$ is totally disconnected, and $mathbb{R}$ is connected and thus has connected image.
Any convergent sequence with limit (like ${frac{1}{n}: n ge 1} cup {0}$) is homeomorphic to $mathbb{N} cup {infty}$, as is easily checked.
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
$endgroup$
– user531587
Dec 5 '18 at 3:55
$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57
add a comment |
$begingroup$
(1) a continuous map $f$ from $mathbb{N} cup {infty}$ into $mathbb{R}$ corresponds to a convergent sequence and its limit, in the sense that for any space $X$, $f: mathbb{N} cup {infty} to X$ is continuous iff $x_n = f(n)$ defines a sequence that converges to $f(infty)$ in $X$. And conversely for every sequence $x_n$ in $X$ that converges to $x$, the function defined by $f(n) = x_n$ for all $n$ and $f(infty) = x$, is continuous from $mathbb{N} cup {infty}$ to $X$.
The other way around (from $mathbb{R}$ to $mathbb{N} cup {infty}$ there are only constant maps as $mathbb{N} cup {infty}$ is totally disconnected, and $mathbb{R}$ is connected and thus has connected image.
Any convergent sequence with limit (like ${frac{1}{n}: n ge 1} cup {0}$) is homeomorphic to $mathbb{N} cup {infty}$, as is easily checked.
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
$endgroup$
– user531587
Dec 5 '18 at 3:55
$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57
add a comment |
$begingroup$
(1) a continuous map $f$ from $mathbb{N} cup {infty}$ into $mathbb{R}$ corresponds to a convergent sequence and its limit, in the sense that for any space $X$, $f: mathbb{N} cup {infty} to X$ is continuous iff $x_n = f(n)$ defines a sequence that converges to $f(infty)$ in $X$. And conversely for every sequence $x_n$ in $X$ that converges to $x$, the function defined by $f(n) = x_n$ for all $n$ and $f(infty) = x$, is continuous from $mathbb{N} cup {infty}$ to $X$.
The other way around (from $mathbb{R}$ to $mathbb{N} cup {infty}$ there are only constant maps as $mathbb{N} cup {infty}$ is totally disconnected, and $mathbb{R}$ is connected and thus has connected image.
Any convergent sequence with limit (like ${frac{1}{n}: n ge 1} cup {0}$) is homeomorphic to $mathbb{N} cup {infty}$, as is easily checked.
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
$endgroup$
(1) a continuous map $f$ from $mathbb{N} cup {infty}$ into $mathbb{R}$ corresponds to a convergent sequence and its limit, in the sense that for any space $X$, $f: mathbb{N} cup {infty} to X$ is continuous iff $x_n = f(n)$ defines a sequence that converges to $f(infty)$ in $X$. And conversely for every sequence $x_n$ in $X$ that converges to $x$, the function defined by $f(n) = x_n$ for all $n$ and $f(infty) = x$, is continuous from $mathbb{N} cup {infty}$ to $X$.
The other way around (from $mathbb{R}$ to $mathbb{N} cup {infty}$ there are only constant maps as $mathbb{N} cup {infty}$ is totally disconnected, and $mathbb{R}$ is connected and thus has connected image.
Any convergent sequence with limit (like ${frac{1}{n}: n ge 1} cup {0}$) is homeomorphic to $mathbb{N} cup {infty}$, as is easily checked.
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
edited Dec 5 '18 at 2:52
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
answered Dec 3 '18 at 6:03
Henno BrandsmaHenno Brandsma
110k347117
110k347117
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
$endgroup$
– user531587
Dec 5 '18 at 3:55
$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57
add a comment |
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
$endgroup$
– user531587
Dec 5 '18 at 3:55
$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
In what sense does a continuous map correspond to a convergent sequence and its limit? What's the precise statement?
$endgroup$
– user531587
Dec 5 '18 at 2:18
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
For the homeomorphism part, I posted a separate question math.stackexchange.com/questions/3026518/…
$endgroup$
– user531587
Dec 5 '18 at 2:34
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
@user531587 I made the statement exact about sequences.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 2:52
$begingroup$
I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
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– user531587
Dec 5 '18 at 3:55
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I think you proved "and conversely" part of the first paragraph in the question referred to above (if $X$ is Hausdorff). I'm not sure how to deal with the other implication. To show that $f(n)$ converges to $f(infty)$, we need to show that any neighborhood of $f(infty)$ contains all $f(n)$ for $n$ large. But we know from continuity that for any nbhd of $f(infty)$ there exists a nbhd of $infty$ whose image lies in the nbhd of $f(infty)$. Any nbhd of $infty$ contains infinitely many elts of $N$, so the nbhd of $f(infty)$ also contains infinitely many pts. Is that how the proof goes?
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– user531587
Dec 5 '18 at 3:55
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@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
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– Henno Brandsma
Dec 5 '18 at 3:57
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@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
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– Henno Brandsma
Dec 5 '18 at 3:57
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Recall that $mathbb{R}$ is connected and the image of a connected space under a continuous map is connected. What you can say then about continuous maps $mathbb{R} to N$? For the other direction, can you extend a function $mathbb{N} to mathbb{R}$ to a continuous map $N to mathbb{R}$? If no, what condition do you need?
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– Luca Carai
Dec 3 '18 at 1:57
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@LucaCarai So from what you said we can conclude that if $mathbb Rto N$ is continuous, then it's image is a connected subset of $N$. I have a conjecture that only singletons are connected subspaces of $N$ (I thought how to prove this, but I'm not sure: this topology is so weird!). If this is so, then all continuous maps are constant. For the other direction, I don't know even what techniques I should use for continuous extensions.
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– user531587
Dec 3 '18 at 2:19
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Go back to the definition of connected sets. Let U = $U_1 sqcup U_2$. What does the topology you have say about U?
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– Joel Pereira
Dec 3 '18 at 2:28
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@JoelPereira That's how I think about it. Both $U_1, U_2$ are open. Either they both do not contain $infty$, or they both have finite complements, or one of the does not contain $infty$ and the other has finite complement. I don't think it says something spacial about $U$ other than $U$ is open...
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– user531587
Dec 3 '18 at 3:08