When teaching someone how to prove a function is uniformly continuous, using epsilon/delta, which example would be among the simplest?
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I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user757
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user757
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
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@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user757
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
calculus limits
edited Mar 3 at 17:30
user757
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
edited Mar 3 at 17:30
user757
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
edited Mar 3 at 17:30
user757
3,1261125
edited Mar 3 at 17:30
user757
3,1261125
edited Mar 3 at 17:30
user757
3,1261125
3,1261125
asked Mar 2 at 22:55
AlecAlec
619311
asked Mar 2 at 22:55
AlecAlec
619311
asked Mar 2 at 22:55
AlecAlec
619311
619311
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
2
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
1
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
1 Answer
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oldest
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I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac1x$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac1x$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac1x$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac1x$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
$endgroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac1x$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15.2k33381
15.2k33381
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
1
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
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2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27