Continuous maps between $mathbb Ncup {infty}$ and $mathbb R$












1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @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
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    @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














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

(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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

(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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

(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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

(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.






share|cite|improve this answer











$endgroup$



(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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 2:52

























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


















  • $begingroup$
    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?
$endgroup$
– user531587
Dec 5 '18 at 3:55




$begingroup$
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




$begingroup$
@user531587 not just infinitely many; all but finitely many which is stronger and gives convergence.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:57


















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