Mixing
Continuity via sequences in general topology [duplicate]
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This question already has an answer here:
Example of topological spaces where sequential continuity does not imply continuity
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I have anexo exercise to answer that if for every sequence $x_{n} rightarrow x$ we have $f(x_{n}) rightarrow f(x)$ than is f continuous or not.
Continuous here means that $f^-1(A)$ is open If $A$ is open
I feel that this is false but can neither proof or show a counter example
PS: Sorry If duplicate but couldnt find the same question on search
general-topology continuity
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
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marked as duplicate by Cheerful Parsnip, jgon, Math1000, Lord_Farin, Henno Brandsma general-topology
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Jan 9 at 22:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Example of topological spaces where sequential continuity does not imply continuity
3 answers
I have anexo exercise to answer that if for every sequence $x_{n} rightarrow x$ we have $f(x_{n}) rightarrow f(x)$ than is f continuous or not.
Continuous here means that $f^-1(A)$ is open If $A$ is open
I feel that this is false but can neither proof or show a counter example
PS: Sorry If duplicate but couldnt find the same question on search
general-topology continuity
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
$endgroup$
marked as duplicate by Cheerful Parsnip, jgon, Math1000, Lord_Farin, Henno Brandsma general-topology
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Jan 9 at 22:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Do you know about uncountable ordinals?
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– Lord Shark the Unknown
Jan 9 at 19:53
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@LordSharktheUnknown never heard of but can look It up
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– Daniel Moraes
Jan 9 at 20:00
add a comment |
$begingroup$
This question already has an answer here:
Example of topological spaces where sequential continuity does not imply continuity
3 answers
I have anexo exercise to answer that if for every sequence $x_{n} rightarrow x$ we have $f(x_{n}) rightarrow f(x)$ than is f continuous or not.
Continuous here means that $f^-1(A)$ is open If $A$ is open
I feel that this is false but can neither proof or show a counter example
PS: Sorry If duplicate but couldnt find the same question on search
general-topology continuity
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
$endgroup$
This question already has an answer here:
Example of topological spaces where sequential continuity does not imply continuity
3 answers
I have anexo exercise to answer that if for every sequence $x_{n} rightarrow x$ we have $f(x_{n}) rightarrow f(x)$ than is f continuous or not.
Continuous here means that $f^-1(A)$ is open If $A$ is open
I feel that this is false but can neither proof or show a counter example
PS: Sorry If duplicate but couldnt find the same question on search
This question already has an answer here:
Example of topological spaces where sequential continuity does not imply continuity
3 answers
general-topology continuity
general-topology continuity
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
asked Jan 9 at 19:52
Daniel MoraesDaniel Moraes
314110
314110
marked as duplicate by Cheerful Parsnip, jgon, Math1000, Lord_Farin, Henno Brandsma general-topology
Users with the general-topology badge can single-handedly close general-topology questions as duplicates and reopen them as needed.
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Jan 9 at 22:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Cheerful Parsnip, jgon, Math1000, Lord_Farin, Henno Brandsma general-topology
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Jan 9 at 22:52
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Do you know about uncountable ordinals?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 19:53
$begingroup$
@LordSharktheUnknown never heard of but can look It up
$endgroup$
– Daniel Moraes
Jan 9 at 20:00
add a comment |
$begingroup$
Do you know about uncountable ordinals?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 19:53
$begingroup$
@LordSharktheUnknown never heard of but can look It up
$endgroup$
– Daniel Moraes
Jan 9 at 20:00
$begingroup$
Do you know about uncountable ordinals?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 19:53
$begingroup$
Do you know about uncountable ordinals?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 19:53
$begingroup$
@LordSharktheUnknown never heard of but can look It up
$endgroup$
– Daniel Moraes
Jan 9 at 20:00
$begingroup$
@LordSharktheUnknown never heard of but can look It up
$endgroup$
– Daniel Moraes
Jan 9 at 20:00
add a comment |
1 Answer
1
active
oldest
votes
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One can only guarantee that a function $f:X to Y$ is continuous $iff$ for every sequence $x_n$ in $X$ converging to $x in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x in X$ has a countable neighbourhood basis. In the comment by $textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
answered Jan 9 at 20:32
greeliousgreelious
465112
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One can only guarantee that a function $f:X to Y$ is continuous $iff$ for every sequence $x_n$ in $X$ converging to $x in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x in X$ has a countable neighbourhood basis. In the comment by $textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
answered Jan 9 at 20:32
greeliousgreelious
465112
$endgroup$
add a comment |
$begingroup$
One can only guarantee that a function $f:X to Y$ is continuous $iff$ for every sequence $x_n$ in $X$ converging to $x in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x in X$ has a countable neighbourhood basis. In the comment by $textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
answered Jan 9 at 20:32
greeliousgreelious
465112
$endgroup$
add a comment |
$begingroup$
One can only guarantee that a function $f:X to Y$ is continuous $iff$ for every sequence $x_n$ in $X$ converging to $x in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x in X$ has a countable neighbourhood basis. In the comment by $textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
answered Jan 9 at 20:32
greeliousgreelious
465112
$endgroup$
One can only guarantee that a function $f:X to Y$ is continuous $iff$ for every sequence $x_n$ in $X$ converging to $x in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x in X$ has a countable neighbourhood basis. In the comment by $textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
answered Jan 9 at 20:32
greeliousgreelious
465112
answered Jan 9 at 20:32
greeliousgreelious
465112
answered Jan 9 at 20:32
greeliousgreelious
465112
answered Jan 9 at 20:32
greeliousgreelious
465112
465112
add a comment |
add a comment |
$begingroup$
Do you know about uncountable ordinals?
$endgroup$
– Lord Shark the Unknown
Jan 9 at 19:53
$begingroup$
@LordSharktheUnknown never heard of but can look It up
$endgroup$
– Daniel Moraes
Jan 9 at 20:00