Prove: Given $x_0$ is a cluster point of a set $S$ and $f:S to mathbb{R}$ then $f$ can have at most one limit...
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Essentially, I need to prove that, given a point $x_0$ in $S$ where $S subseteq mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.
This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.
Cluster point: if $x_0$ is a cluster point, then
$forall epsilon gt 0$, $(x_0- epsilon , x_0+ epsilon ) cap (S setminus {x_0}) neq phi $
definition of continuity: $forall epsilon gt 0$, $exists delta gt 0$ such that $0 lt |x-x_0| lt delta$, $x in S$, implies $|f(x)-f(x_0)| lt epsilon$.
I'm not too sure how I would approach this!
real-analysis proof-writing
asked Nov 25 at 22:53
Bob
404
add a comment |
up vote
0
down vote
favorite
Essentially, I need to prove that, given a point $x_0$ in $S$ where $S subseteq mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.
This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.
Cluster point: if $x_0$ is a cluster point, then
$forall epsilon gt 0$, $(x_0- epsilon , x_0+ epsilon ) cap (S setminus {x_0}) neq phi $
definition of continuity: $forall epsilon gt 0$, $exists delta gt 0$ such that $0 lt |x-x_0| lt delta$, $x in S$, implies $|f(x)-f(x_0)| lt epsilon$.
I'm not too sure how I would approach this!
real-analysis proof-writing
asked Nov 25 at 22:53
Bob
404
Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Essentially, I need to prove that, given a point $x_0$ in $S$ where $S subseteq mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.
This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.
Cluster point: if $x_0$ is a cluster point, then
$forall epsilon gt 0$, $(x_0- epsilon , x_0+ epsilon ) cap (S setminus {x_0}) neq phi $
definition of continuity: $forall epsilon gt 0$, $exists delta gt 0$ such that $0 lt |x-x_0| lt delta$, $x in S$, implies $|f(x)-f(x_0)| lt epsilon$.
I'm not too sure how I would approach this!
real-analysis proof-writing
asked Nov 25 at 22:53
Bob
404
Essentially, I need to prove that, given a point $x_0$ in $S$ where $S subseteq mathbb{R}$, as the $x$ value converges to $x_0$, $f(x)$ converges to only one $f(x_0)$. This is poking at the idea that for every input, $x$, there cannot be more than one output, $f(x)$.
This is something that most of us knew from algebra 1, but I need to prove this statement using the definition of cluster points, continuity, etc.
Cluster point: if $x_0$ is a cluster point, then
$forall epsilon gt 0$, $(x_0- epsilon , x_0+ epsilon ) cap (S setminus {x_0}) neq phi $
definition of continuity: $forall epsilon gt 0$, $exists delta gt 0$ such that $0 lt |x-x_0| lt delta$, $x in S$, implies $|f(x)-f(x_0)| lt epsilon$.
I'm not too sure how I would approach this!
real-analysis proof-writing
real-analysis proof-writing
asked Nov 25 at 22:53
Bob
404
asked Nov 25 at 22:53
Bob
404
asked Nov 25 at 22:53
Bob
404
asked Nov 25 at 22:53
Bob
404
asked Nov 25 at 22:53
Bob
404
404
Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35
add a comment |
Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35
Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35
Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35
add a comment |
2 Answers
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0
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First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.
Now to show that $f$ has at most one limit as $x to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).
In other words, we can assume that the limit exists, and let
$$ lim_{x to x_0} f(x) = l_1$$ and $$ lim_{x to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.
Given $epsilon > 0$, $exists delta_1 > 0$ such that
$$|f(x)-l_1| < frac{epsilon}{2}, forall x in V_{delta_1}(x_0)cap{S}setminus{{x_0
}}$$
Also, $exists delta_2>0$ such that
$$|f(x)-l_2|<frac{epsilon}{2}, forall x in V_{delta_2}(x_0)cap{S}setminus{{x_0}}$$
. Now take $delta = min{delta_1, delta_2}$.
$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| le |l_1-f(x)|+|f(x)-l_2| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon, forall x in V_delta(x_0) cap Ssetminus {{x_0}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $forall epsilon > 0, 0le a < epsilon$, can you prove by contradiction that $a = 0$?)
answered Nov 26 at 4:49
tonychow0929
24122
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0
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Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $left| a-b right|$? And how might this help you use the definition of a limit?
answered Nov 26 at 3:18
nick
353
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.
Now to show that $f$ has at most one limit as $x to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).
In other words, we can assume that the limit exists, and let
$$ lim_{x to x_0} f(x) = l_1$$ and $$ lim_{x to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.
Given $epsilon > 0$, $exists delta_1 > 0$ such that
$$|f(x)-l_1| < frac{epsilon}{2}, forall x in V_{delta_1}(x_0)cap{S}setminus{{x_0
}}$$
Also, $exists delta_2>0$ such that
$$|f(x)-l_2|<frac{epsilon}{2}, forall x in V_{delta_2}(x_0)cap{S}setminus{{x_0}}$$
. Now take $delta = min{delta_1, delta_2}$.
$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| le |l_1-f(x)|+|f(x)-l_2| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon, forall x in V_delta(x_0) cap Ssetminus {{x_0}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $forall epsilon > 0, 0le a < epsilon$, can you prove by contradiction that $a = 0$?)
answered Nov 26 at 4:49
tonychow0929
24122
add a comment |
up vote
0
down vote
accepted
First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.
Now to show that $f$ has at most one limit as $x to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).
In other words, we can assume that the limit exists, and let
$$ lim_{x to x_0} f(x) = l_1$$ and $$ lim_{x to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.
Given $epsilon > 0$, $exists delta_1 > 0$ such that
$$|f(x)-l_1| < frac{epsilon}{2}, forall x in V_{delta_1}(x_0)cap{S}setminus{{x_0
}}$$
Also, $exists delta_2>0$ such that
$$|f(x)-l_2|<frac{epsilon}{2}, forall x in V_{delta_2}(x_0)cap{S}setminus{{x_0}}$$
. Now take $delta = min{delta_1, delta_2}$.
$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| le |l_1-f(x)|+|f(x)-l_2| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon, forall x in V_delta(x_0) cap Ssetminus {{x_0}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $forall epsilon > 0, 0le a < epsilon$, can you prove by contradiction that $a = 0$?)
answered Nov 26 at 4:49
tonychow0929
24122
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.
Now to show that $f$ has at most one limit as $x to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).
In other words, we can assume that the limit exists, and let
$$ lim_{x to x_0} f(x) = l_1$$ and $$ lim_{x to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.
Given $epsilon > 0$, $exists delta_1 > 0$ such that
$$|f(x)-l_1| < frac{epsilon}{2}, forall x in V_{delta_1}(x_0)cap{S}setminus{{x_0
}}$$
Also, $exists delta_2>0$ such that
$$|f(x)-l_2|<frac{epsilon}{2}, forall x in V_{delta_2}(x_0)cap{S}setminus{{x_0}}$$
. Now take $delta = min{delta_1, delta_2}$.
$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| le |l_1-f(x)|+|f(x)-l_2| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon, forall x in V_delta(x_0) cap Ssetminus {{x_0}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $forall epsilon > 0, 0le a < epsilon$, can you prove by contradiction that $a = 0$?)
answered Nov 26 at 4:49
tonychow0929
24122
First, the existence of limit at a point (say, $x_0$) does not depend on the value of the function at $x_0$. This is important because, without this idea, limit and continuity become more or less useless.
Now to show that $f$ has at most one limit as $x to x_0$, we need to show that either the limit does not exist ("zero" limit), or the limit is unique if it exists ("one" limit).
In other words, we can assume that the limit exists, and let
$$ lim_{x to x_0} f(x) = l_1$$ and $$ lim_{x to x_0} f(x) = l_2$$. Now we want to show that $l_1 = l_2$. Start from the definition.
Given $epsilon > 0$, $exists delta_1 > 0$ such that
$$|f(x)-l_1| < frac{epsilon}{2}, forall x in V_{delta_1}(x_0)cap{S}setminus{{x_0
}}$$
Also, $exists delta_2>0$ such that
$$|f(x)-l_2|<frac{epsilon}{2}, forall x in V_{delta_2}(x_0)cap{S}setminus{{x_0}}$$
. Now take $delta = min{delta_1, delta_2}$.
$$|l_1-l_2| = |l_1-f(x)+f(x)-l_2| le |l_1-f(x)|+|f(x)-l_2| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon, forall x in V_delta(x_0) cap Ssetminus {{x_0}}$$. Since $x_0$ is a cluster point, $|l_1-l_2|$ is arbitrary small (but still non-negative) on a non-empty set. Can you conclude from here? (An extra hint: Suppose $forall epsilon > 0, 0le a < epsilon$, can you prove by contradiction that $a = 0$?)
answered Nov 26 at 4:49
tonychow0929
24122
answered Nov 26 at 4:49
tonychow0929
24122
answered Nov 26 at 4:49
tonychow0929
24122
answered Nov 26 at 4:49
tonychow0929
24122
24122
add a comment |
add a comment |
up vote
0
down vote
Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $left| a-b right|$? And how might this help you use the definition of a limit?
answered Nov 26 at 3:18
nick
353
add a comment |
up vote
0
down vote
Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $left| a-b right|$? And how might this help you use the definition of a limit?
answered Nov 26 at 3:18
nick
353
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $left| a-b right|$? And how might this help you use the definition of a limit?
answered Nov 26 at 3:18
nick
353
Hint: A really classic way to approach uniqueness questions in analysis is by contradiction. Suppose $f$ had more than one limit (ie, two - say $a$ and $b$). What do you know about $left| a-b right|$? And how might this help you use the definition of a limit?
answered Nov 26 at 3:18
nick
353
answered Nov 26 at 3:18
nick
353
answered Nov 26 at 3:18
nick
353
answered Nov 26 at 3:18
nick
353
353
add a comment |
add a comment |
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Since in Hausdorff spaces limits are unique, there is nothing more to consider. Is f supposed to be continuous?
– William Elliot
Nov 26 at 1:35