Intermediate value property not implies continuous - Revision history

Vipul: /* Proof */

2021-09-24T16:17:48Z

Proof

← Older revision		Revision as of 16:17, 24 September 2021
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	[[File:Sin1byxzoomin2.png\|600px]]		[[File:Sin1byxzoomin2.png\|600px]]

	=== No limit at zero, hence discontinuous at zero ===		=== No limit at zero, hence discontinuous at zero (continuous elsewhere) ===

	The function takes the value 1 at all points of the form <math>2/((4n + 1)\pi)</math> for integers <math>n</math>, and takes the value -1 at all points of the form <math>2/((4n - 1)\pi)</math> for integers <math>n</math>. We can use this to see that the function takes the values 1 and -1 for points arbitrarily close to 0. It therefore cannot have a limit at 0. (In the <math>\epsilon-\delta</math> definition, picking any <math>\epsilon \le 1</math> will yield trouble because no matter how small the <math>\delta</math> we pick, the value of the function cannot be trapped in an interval of size <math>\epsilon</math> because it must visit both 1 and -1).		The function takes the value 1 at all points of the form <math>2/((4n + 1)\pi)</math> for integers <math>n</math>, and takes the value -1 at all points of the form <math>2/((4n - 1)\pi)</math> for integers <math>n</math>. We can use this to see that the function takes the values 1 and -1 for points arbitrarily close to 0. It therefore cannot have a limit at 0. (In the <math>\epsilon-\delta</math> definition, picking any <math>\epsilon \le 1</math> will yield trouble because no matter how small the <math>\delta</math> we pick, the value of the function cannot be trapped in an interval of size <math>\epsilon</math> because it must visit both 1 and -1).

Vipul: /* Example of the sine of reciprocal function */

2021-09-24T16:07:45Z

Example of the sine of reciprocal function

← Older revision		Revision as of 16:07, 24 September 2021
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	== Proof ==		== Proof ==

	=== Example of the sine of reciprocal function ===		=== Example of the sine of reciprocal function (isolated discontinuity) ===

	{{further\|[[sine of reciprocal function]]}}		{{further\|[[sine of reciprocal function]]}}

Vipul: /* Example of the sine of reciprocal function */

2021-09-24T16:07:09Z

Example of the sine of reciprocal function

← Older revision		Revision as of 16:07, 24 September 2021
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	NOTE: For the value at zero, we could choose any value in <math>[-1,1]</math> instead of 0 and all the arguments below would work.		NOTE: For the value at zero, we could choose any value in <math>[-1,1]</math> instead of 0 and all the arguments below would work.

			Here is the graph, at various levels of magnification:

			[[File:Sin1byx.png\|600px]]

			[[File:Sin1byx-1to1.png\|600px]]

			[[File:Sin1byxzoomin1.png\|600px]]

			[[File:Sin1byxzoomin2.png\|600px]]

	=== No limit at zero, hence discontinuous at zero ===		=== No limit at zero, hence discontinuous at zero ===

Vipul: /* Related facts */

2021-09-24T16:06:48Z

Related facts

← Older revision		Revision as of 16:06, 24 September 2021
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	* [[Continuous not implies differentiable]]		* [[Continuous not implies differentiable]]
	* [[Derivative of differentiable function not ~~implies~~ continuous]] (combined with [[derivative of differentiable function satisfies intermediate value property]], this would imply the statement of the current page)		* [[Derivative of differentiable function need not be continuous]] (combined with [[derivative of differentiable function satisfies intermediate value property]], this would imply the statement of the current page)

	== Proof ==		== Proof ==

Vipul at 16:06, 24 September 2021

2021-09-24T16:06:34Z

@@ Line 7: / Line 7: @@
 It is possible to have a real-valued function defined on all real numbers that satisfies the [[intermediate value property]] but that is not a [[continuous function]], i.e., there is at least one point where it is not continuous.
 == Proof ==
@@ Line 27: / Line 32: @@
 As noted above, the function takes values of 1 and -1 arbitrarily close to 0. It is also continuous on the right of 0 and on the left of 0. It therefore satisfies the intermediate value property on either side of 0, and in particular, takes all values in the interval <math>[-1,1]</math> arbitrarily close to zero on either side. Also, the range of the function is <math>[-1,1]</math>, so basically the function takes its full range on inputs arbitrarily close to zero. Based on this, we see that the function satisfies the intermediate value property between 0 and any point, and also across zero (between a positive and negative value). Since it already satisfies the intermediate value property on either side of zero, it satisfies the intermediate value property globally.

Vipul: /* = No limit at zero, hence discontinuous at zero */

2021-09-24T15:58:42Z

= No limit at zero, hence discontinuous at zero

← Older revision		Revision as of 15:58, 24 September 2021
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	NOTE: For the value at zero, we could choose any value in <math>[-1,1]</math> instead of 0 and all the arguments below would work.		NOTE: For the value at zero, we could choose any value in <math>[-1,1]</math> instead of 0 and all the arguments below would work.

	==== No limit at zero, hence discontinuous at zero ===		=== No limit at zero, hence discontinuous at zero ===

	The function takes the value 1 at all points of the form <math>2/((4n + 1)\pi)</math> for integers <math>n</math>, and takes the value -1 at all points of the form <math>2/((4n - 1)\pi)</math> for integers <math>n</math>. We can use this to see that the function takes the values 1 and -1 for points arbitrarily close to 0. It therefore cannot have a limit at 0. (In the <math>\epsilon-\delta</math> definition, picking any <math>\epsilon \le 1</math> will yield trouble because no matter how small the <math>\delta</math> we pick, the value of the function cannot be trapped in an interval of size <math>\epsilon</math> because it must visit both 1 and -1).		The function takes the value 1 at all points of the form <math>2/((4n + 1)\pi)</math> for integers <math>n</math>, and takes the value -1 at all points of the form <math>2/((4n - 1)\pi)</math> for integers <math>n</math>. We can use this to see that the function takes the values 1 and -1 for points arbitrarily close to 0. It therefore cannot have a limit at 0. (In the <math>\epsilon-\delta</math> definition, picking any <math>\epsilon \le 1</math> will yield trouble because no matter how small the <math>\delta</math> we pick, the value of the function cannot be trapped in an interval of size <math>\epsilon</math> because it must visit both 1 and -1).

Vipul: /* Example of the sine of reciprocal function */

2021-09-24T15:56:28Z

Example of the sine of reciprocal function

← Older revision		Revision as of 15:56, 24 September 2021
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	<math>f(x) := \left\lbrace \begin{array}{rl} \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{array}\right.</math>		<math>f(x) := \left\lbrace \begin{array}{rl} \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{array}\right.</math>

			NOTE: For the value at zero, we could choose any value in <math>[-1,1]</math> instead of 0 and all the arguments below would work.

	==== No limit at zero, hence discontinuous at zero ===		==== No limit at zero, hence discontinuous at zero ===

Vipul: Created page with "{{function property non-implication |stronger = intermediate value property |weaker = continuous function }} == Definition == It is possible to have a real-valued function d..."

2021-09-24T15:55:42Z

Created page with "{{function property non-implication |stronger = intermediate value property |weaker = continuous function }} == Definition == It is possible to have a real-valued function d..."

New page

{{function property non-implication
|stronger = intermediate value property
|weaker = continuous function
}}

== Definition ==

It is possible to have a real-valued function defined on all real numbers that satisfies the [[intermediate value property]] but that is not a [[continuous function]], i.e., there is at least one point where it is not continuous.

== Proof ==

=== Example of the sine of reciprocal function ===

{{further|[[sine of reciprocal function]]}}

Consider the function that equals the [[sine of reciprocal function]] for nonzero values and is zero at zero. Explicitly:

<math>f(x) := \left\lbrace \begin{array}{rl} \sin(1/x), & x \ne 0 \\ 0, & x = 0 \end{array}\right.</math>

==== No limit at zero, hence discontinuous at zero ===

The function takes the value 1 at all points of the form <math>2/((4n + 1)\pi)</math> for integers <math>n</math>, and takes the value -1 at all points of the form <math>2/((4n - 1)\pi)</math> for integers <math>n</math>. We can use this to see that the function takes the values 1 and -1 for points arbitrarily close to 0. It therefore cannot have a limit at 0. (In the <math>\epsilon-\delta</math> definition, picking any <math>\epsilon \le 1</math> will yield trouble because no matter how small the <math>\delta</math> we pick, the value of the function cannot be trapped in an interval of size <math>\epsilon</math> because it must visit both 1 and -1).

==== Intermediate value property held everywhere ====

As noted above, the function takes values of 1 and -1 arbitrarily close to 0. It is also continuous on the right of 0 and on the left of 0. It therefore satisfies the intermediate value property on either side of 0, and in particular, takes all values in the interval <math>[-1,1]</math> arbitrarily close to zero on either side. Also, the range of the function is <math>[-1,1]</math>, so basically the function takes its full range on inputs arbitrarily close to zero. Based on this, we see that the function satisfies the intermediate value property between 0 and any point, and also across zero (between a positive and negative value). Since it already satisfies the intermediate value property on either side of zero, it satisfies the intermediate value property globally.