Proving a function is continuous but unbounded

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Problem:

Hi, I'm working on the following math problem:

Suppose a set $S$ contains a sequence that converges to a point
$x_0$ that is not in $S$. Show that the function $f : S
rightarrow mathbbR$ defined by $f(x) = 1/|x - x_0|$ for all $x$
in $S$ is continuous but unbounded.

Here is the definition of continuity that I am referring to. Note, however, that I am not using the typical $epsilon-delta$ definition of continuity:

Definition: A function $f : D rightarrow mathbbR$ is said to be continuous at the point $x_0$ in $D$ provided that whenever
$x_n$ is a sequence in $D$ that converges to $x_0$, the image
sequence $f(x_n)$ converges to $f(x_0)$.

My attempt:

Let $x_n$ be a sequence in $S$ that converges to the point $x_0$, which is not in $S$. By the definition of convergence provided, this means $forall epsilon > 0, exists N$ s.t.

$$|x_n - x_0| < epsilon$$

$forall n geq N$. To prove continuity at $x_0$, I must show that there is some index $N'$ so that $forall epsilon > 0$,

$$|f(x_n) - f(x_0) | < epsilon$$

$forall epsilon > 0$. By our definition of $f$, I can write

$$|f(x_n) - f(x_0) | = |1/(x - x_0) - 1/(x_0 - x_0)|,$$

but the right-hand term is not defined, so I'm not sure if my approach is correct.

Some additional info:

In case it helps, this problem is part of a two-part problem. The first part of this problem asked me to show that if $S$ contains an unbounded sequence, then show that the function $f : S rightarrow mathbbR$ defined by $f(x) = x$ for all $x$ in $S$ is continuous but unbounded. So, I can cite this fact if I want. I'm guessing that I'll have to since it's part of the same problem, but I'm not sure where that'll come into play.

Thank you to anyone who tries to help.

asked 1 hour ago

stackofhay42

373213

add a commentÂ |Â

up vote
2
down vote

favorite

Problem:

Hi, I'm working on the following math problem:

Suppose a set $S$ contains a sequence that converges to a point
$x_0$ that is not in $S$. Show that the function $f : S
rightarrow mathbbR$ defined by $f(x) = 1/|x - x_0|$ for all $x$
in $S$ is continuous but unbounded.

Here is the definition of continuity that I am referring to. Note, however, that I am not using the typical $epsilon-delta$ definition of continuity:

Definition: A function $f : D rightarrow mathbbR$ is said to be continuous at the point $x_0$ in $D$ provided that whenever
$x_n$ is a sequence in $D$ that converges to $x_0$, the image
sequence $f(x_n)$ converges to $f(x_0)$.

My attempt:

Let $x_n$ be a sequence in $S$ that converges to the point $x_0$, which is not in $S$. By the definition of convergence provided, this means $forall epsilon > 0, exists N$ s.t.

$$|x_n - x_0| < epsilon$$

$forall n geq N$. To prove continuity at $x_0$, I must show that there is some index $N'$ so that $forall epsilon > 0$,

$$|f(x_n) - f(x_0) | < epsilon$$

$forall epsilon > 0$. By our definition of $f$, I can write

$$|f(x_n) - f(x_0) | = |1/(x - x_0) - 1/(x_0 - x_0)|,$$

but the right-hand term is not defined, so I'm not sure if my approach is correct.

Some additional info:

Thank you to anyone who tries to help.

asked 1 hour ago

stackofhay42

373213

add a commentÂ |Â

up vote
2
down vote

favorite

Problem:

Hi, I'm working on the following math problem:

Suppose a set $S$ contains a sequence that converges to a point
$x_0$ that is not in $S$. Show that the function $f : S
rightarrow mathbbR$ defined by $f(x) = 1/|x - x_0|$ for all $x$
in $S$ is continuous but unbounded.

Here is the definition of continuity that I am referring to. Note, however, that I am not using the typical $epsilon-delta$ definition of continuity:

Definition: A function $f : D rightarrow mathbbR$ is said to be continuous at the point $x_0$ in $D$ provided that whenever
$x_n$ is a sequence in $D$ that converges to $x_0$, the image
sequence $f(x_n)$ converges to $f(x_0)$.

My attempt:

Let $x_n$ be a sequence in $S$ that converges to the point $x_0$, which is not in $S$. By the definition of convergence provided, this means $forall epsilon > 0, exists N$ s.t.

$$|x_n - x_0| < epsilon$$

$forall n geq N$. To prove continuity at $x_0$, I must show that there is some index $N'$ so that $forall epsilon > 0$,

$$|f(x_n) - f(x_0) | < epsilon$$

$forall epsilon > 0$. By our definition of $f$, I can write

$$|f(x_n) - f(x_0) | = |1/(x - x_0) - 1/(x_0 - x_0)|,$$

but the right-hand term is not defined, so I'm not sure if my approach is correct.

Some additional info:

Thank you to anyone who tries to help.

asked 1 hour ago

stackofhay42

373213

Problem:

Hi, I'm working on the following math problem:

Suppose a set $S$ contains a sequence that converges to a point
$x_0$ that is not in $S$. Show that the function $f : S
rightarrow mathbbR$ defined by $f(x) = 1/|x - x_0|$ for all $x$
in $S$ is continuous but unbounded.

Here is the definition of continuity that I am referring to. Note, however, that I am not using the typical $epsilon-delta$ definition of continuity:

Definition: A function $f : D rightarrow mathbbR$ is said to be continuous at the point $x_0$ in $D$ provided that whenever
$x_n$ is a sequence in $D$ that converges to $x_0$, the image
sequence $f(x_n)$ converges to $f(x_0)$.

My attempt:

Let $x_n$ be a sequence in $S$ that converges to the point $x_0$, which is not in $S$. By the definition of convergence provided, this means $forall epsilon > 0, exists N$ s.t.

$$|x_n - x_0| < epsilon$$

$forall n geq N$. To prove continuity at $x_0$, I must show that there is some index $N'$ so that $forall epsilon > 0$,

$$|f(x_n) - f(x_0) | < epsilon$$

$forall epsilon > 0$. By our definition of $f$, I can write

$$|f(x_n) - f(x_0) | = |1/(x - x_0) - 1/(x_0 - x_0)|,$$

but the right-hand term is not defined, so I'm not sure if my approach is correct.

Some additional info:

Thank you to anyone who tries to help.

real-analysis convergence continuity

asked 1 hour ago

stackofhay42

373213

asked 1 hour ago

stackofhay42

373213

asked 1 hour ago

stackofhay42

373213

asked 1 hour ago

stackofhay42

373213

asked 1 hour ago

stackofhay42

373213

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
3
down vote

Your approach is not correct because $x_0notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

the function $x$ is continuous;

so, $x-x_0$ is continuous;

since the absolute value function is continuous, $lvert x-x_0rvert$ is continuous;

therefore, $dfrac1lvert x-x_0rvert$ is continuous.

And $f$ is unbounded because, if $(x_n)_ninmathbb N$ is such that $lim_ntoinftyx_n=x_0$, then $lim_ntoinftyf(x_n)=+infty$.

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

add a commentÂ |Â

up vote
2
down vote

The problem asks to show continuity for all $x in S$ and $x_0notin S$!

P.S. Note that
$$left|frac1 - frac1x - x_0right|=frac\
leq frac$$
because, by triangle inequality, $||a|-|b||leq |a-b|$.

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Â |Â
show 1 more comment

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

Your approach is not correct because $x_0notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

the function $x$ is continuous;

so, $x-x_0$ is continuous;

since the absolute value function is continuous, $lvert x-x_0rvert$ is continuous;

therefore, $dfrac1lvert x-x_0rvert$ is continuous.

And $f$ is unbounded because, if $(x_n)_ninmathbb N$ is such that $lim_ntoinftyx_n=x_0$, then $lim_ntoinftyf(x_n)=+infty$.

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

add a commentÂ |Â

up vote
3
down vote

Your approach is not correct because $x_0notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

the function $x$ is continuous;

so, $x-x_0$ is continuous;

since the absolute value function is continuous, $lvert x-x_0rvert$ is continuous;

therefore, $dfrac1lvert x-x_0rvert$ is continuous.

And $f$ is unbounded because, if $(x_n)_ninmathbb N$ is such that $lim_ntoinftyx_n=x_0$, then $lim_ntoinftyf(x_n)=+infty$.

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

add a commentÂ |Â

up vote
3
down vote

Your approach is not correct because $x_0notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

the function $x$ is continuous;

so, $x-x_0$ is continuous;

since the absolute value function is continuous, $lvert x-x_0rvert$ is continuous;

therefore, $dfrac1lvert x-x_0rvert$ is continuous.

And $f$ is unbounded because, if $(x_n)_ninmathbb N$ is such that $lim_ntoinftyx_n=x_0$, then $lim_ntoinftyf(x_n)=+infty$.

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

Your approach is not correct because $x_0notin D$. So, it makes no sense to ask whethr $f$ is continuous at $x_0$. The function $f$ is continuous because, for instance:

the function $x$ is continuous;

so, $x-x_0$ is continuous;

since the absolute value function is continuous, $lvert x-x_0rvert$ is continuous;

therefore, $dfrac1lvert x-x_0rvert$ is continuous.

And $f$ is unbounded because, if $(x_n)_ninmathbb N$ is such that $lim_ntoinftyx_n=x_0$, then $lim_ntoinftyf(x_n)=+infty$.

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

edited 20 mins ago

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

answered 1 hour ago

JosÃ© Carlos Santos

129k17103191

answered 1 hour ago

129k17103191

add a commentÂ |Â

up vote
2
down vote

The problem asks to show continuity for all $x in S$ and $x_0notin S$!

P.S. Note that
$$left|frac1 - frac1x - x_0right|=frac\
leq frac$$
because, by triangle inequality, $||a|-|b||leq |a-b|$.

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Â |Â
show 1 more comment

up vote
2
down vote

The problem asks to show continuity for all $x in S$ and $x_0notin S$!

P.S. Note that
$$left|frac1 - frac1x - x_0right|=frac\
leq frac$$
because, by triangle inequality, $||a|-|b||leq |a-b|$.

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Â |Â
show 1 more comment

up vote
2
down vote

The problem asks to show continuity for all $x in S$ and $x_0notin S$!

P.S. Note that
$$left|frac1 - frac1x - x_0right|=frac\
leq frac$$
because, by triangle inequality, $||a|-|b||leq |a-b|$.

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

The problem asks to show continuity for all $x in S$ and $x_0notin S$!

P.S. Note that
$$left|frac1 - frac1x - x_0right|=frac\
leq frac$$
because, by triangle inequality, $||a|-|b||leq |a-b|$.

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

edited 53 mins ago

answered 1 hour ago

Robert Z

86.9k1056127

answered 1 hour ago

Robert Z

86.9k1056127

answered 1 hour ago

Robert Z

86.9k1056127

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Â |Â
show 1 more comment

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Thanks, this helps clarify things. I cannot figure out why the first equality (that is, $|f(x_n) - f(x)| < epsilon$) is true. Using my definition of convergence, I think I'm supposed to choose a sequence $x_n$ in $S$ that converges to $x$. This means that $forall epsilon > 0$, there exists an index $N$ such that $|x_n - x| < epsilon$ for all $n geq N$. Then I believe I need to use this fact to show $|f(x_n) - f(x)| < epsilon$. I tried using reverse triangle inequality on the expression for $|f(x_n) - f(x)|$, but I didn't get anywhere.
â€“Â stackofhay42
1 hour ago

Use different $epsilon$s: you know that for $epsilon'>0$, $|x_nÃ¢ÂˆÂ’x|<epsilon'$ for all $nÃ¢Â‰Â¥N$. For $epsilon>0$, show that $|f(x_n)Ã¢ÂˆÂ’f(x)|<ÃÂµ$ for all $nÃ¢Â‰Â¥N$.
â€“Â Robert Z
1 hour ago

@stackofhay42 See my P.S.
â€“Â Robert Z
1 hour ago

hey, I have been trying for the last half hour and I still cannot get anything from your edit.
â€“Â stackofhay42
42 mins ago

@stackofhay42 If $x_nto x$ the RHS in the P.S. goes to zero.
â€“Â Robert Z
33 mins ago

Â |Â
show 1 more comment

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