simple question on internal.

Thread Starter

MichealY

Joined Apr 9, 2009
49
Is closed interval bounded interval?
If so,why in Numerical Analysis 7e by Burden,there is a statement said'on a closed and bounded interval'?

Thanks.
MichealY.
 
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studiot

Joined Nov 9, 2007
4,998
I think the book has slightly loose terminology.

An interval may be closed in which case it includes its endpoints

or open in which case it does not.

It may be open at one end and closed at the other or open or closed at both ends.

Note all this refers to the x axis or independant variable

Similarly the interval may be bounded above or bounded below or both or neither. This refers to the y axis or dependant variable.

For example the function y = 1/x is unbounded above as y tends to infinity as x gets smaller. If we limit ourselves to positive values of x then the function is bounded below as it never gets below zero, so zero is a lower bound.

On the x axis the interval is open as x can increase without limit towards infinity but it never gets there as infinity is not a number.
The other end of the interval is also open as x can never actually each zero either.

On the other hand the set of positive integers less than 10 is bounded above and below and the interval is closed at both ends, since x=1 and x=9 are both members of the set.

Closed intervals or sets are often written like this [1,2,3,4,5,6,7,8,9]

and open intervals with the brackets the other way round ]y=1/x: x\(\geq\)0[
 
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Thread Starter

MichealY

Joined Apr 9, 2009
49
PapaBravo,

Thanks.But it is very hard for me to recieve this definition.

Because,for six years,I have never been told that (-∞,b] or [a,+∞) are closed intervals.The textbook and teacher called them infinite intervals.
The definition is always be "A set of number consisting all the numbers of
between a give pair of numbers and include endpoints".So,are +∞ or -∞
endpoints?

But if closed interval is an interval that include all its limit points,then (-∞,+∞),[a,+∞) and (-∞,b] are all closed intervals.Strange but "a closed and bounded" could be explained,because closed intervals could be unbounded.Also,"closed and bounded intervals" means intervals as [a,b].Isn't it?

studiot,

Thanks,but I think you mistook my words.It is the interval that bounds not the function itself bounds.So both closed and bounded only apply to the x-axis or independent variable.

Here is the orginal statement from the textbook,
"Given any function,defined and continuous on a closed and bounded interval,there exisits a polynomial that is as "close" to the given function as desired.".
 

studiot

Joined Nov 9, 2007
4,998
It is the interval that bounds not the function itself bounds
Not exactly. The terms bounded, open, closed, maximum, minimum, supremum, infimum are actually from set theory and can be applied to any set of numbers. An interval is a particular type of set of numbers. It is merely a chunk of one axis. A function can certainly be bounded or unbounded. Tanx is unbounded but sinx is bounded.

However I owe you an apology. This bit I posted is incorrect and should be ignored. I have removed the original.

It is an important theorem that a set which is bounded above has a greatest member and a set which is bounded below has a least member.
In order to help clarify I am attaching an extract from a really good and easy to understand book that I recommend. He uses curved brackets for open intervals rather than reversed ones.

Mathematical Analysis a straightforward approach

K. G. Binmore

Cambridge University Press
 

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Thread Starter

MichealY

Joined Apr 9, 2009
49
Not exactly. The terms bounded, open, closed, maximum, minimum, supremum, infimum are actually from set theory and can be applied to any set of numbers. An interval is a particular type of set of numbers. It is merely a chunk of one axis. A function can certainly be bounded or unbounded. Tanx is unbounded but sinx is bounded.
Exactly.But in this statement in Numerical Analysis,it is the domain interval that bounds not the range of the function bounds.

It is an important theorem that a set which is bounded above has a greatest member and a set which is bounded below has a least member.
Ashamed.I have not recognize it,but it should be true to bounded intervals(or say,functions that is contiunous on bounded range).

I think the book has slightly loose terminology.

An interval may be closed in which case it includes its endpoints

or open in which case it does not.
As PapaBravo says,closed intervals aren't bounded intervals.So the statement and the textbook are not lossing terminology.

Moreover,according the reference he gave,closed intervals is defined as "intervals that get all of its limit point".So if +∞,-∞ are endpoints,then the definition you give is right,otherwise it is not.But,I do not think +∞,or -∞ are endpoints actually.

PS:Thanks for your papers,it helps.
 

studiot

Joined Nov 9, 2007
4,998
Sorry I can't agree with the definition in PapaBravo's link.

infinity (and don't forget there is more than one infinity) is not a 'point' of any sort, nor is it a member of any set of numbers.

Therefore it is not a 'limit point'.

The term is not necessary and I try to avoid it.

You get in to all sorts of self consistency problems if you try to treat infinity as a point.

An interval and a domain are different animals. An interval is just that and can stand alone without a function.

For example the closed interval [1,2] includes its endpoints 1 and 2
Whilst the open interval ]1,2[ does not.

It is correct to talk of the open interval ]-∞, 0[
or the half open interval ]-∞, 0]
but it is not correct to talk of the closed interval [-∞, 0]

It is important to realize that any old set of numbers is not an interval, an interval includes all the real numbers between the endpoints. You cannot choose just some of them.

So the set {1,3,5,7, 9.......} although infinite is not an interval. It, or a subset, could however be a domain for a function an be bounded or unbounded, open or closed.

As such all intervals are infinite sets. A domain can be finite or infinite.

The term domain only makes sense in the context of a function with a range (also called the co-domain)

So the statement in your book 'domain interval' is consistent and carries the useful information that whatever functions there are apply to all the (infinite) possible numbers in the domain because the domain is also an interval.

The Question of boundedness and openness applies to both the domain and the function, even though you book may have chosen to concentrate on only one of these. Read further, I suspect you will find the other covered later.
 
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Thread Starter

MichealY

Joined Apr 9, 2009
49
to studiot,
According to the papers you gave,intervals [a,b],[a,+∞) and (-∞,b] are closed.So interval of (-∞,0] is closed.(In the ISO notation,does ]-∞,0] equal to [-∞,0]?)

According to the wikipedia:http://en.wikipedia.org/wiki/Interval_(mathematics),there is a terminology saying 'Infinite endpoints'.

"Given any function,defined and continuous on a closed and bounded interval,there exisits a polynomial that is as "close" to the given function as desired.".
The orignal statement shows that the interval is for domain.

MichealY.
 

studiot

Joined Nov 9, 2007
4,998
According to the papers you gave,intervals [a,b],[a,+∞) and (-∞,b] are closed.So interval of (-∞,0] is closed.(In the ISO notation,does ]-∞,0] equal to [-∞,0]?)
Good question, you were quick to spot the discrepancy.

Note that Binmore avoids writing [-∞ anywhere. He only puts open brackets (-∞ around infinities.
I did say that people get into difficulties when employing ∞. I have not seen any author use the closed bracket around infinities.

The difficulty is that the interval (-∞, +∞) or ]-∞, +∞[ is regarded as both open and closed, but never written [-∞, +∞]

The intervals (-∞, a) and (a, +∞) are generally regarded as closed (as Binmore says) but written as if they were open at the infinity end.

Correction
As pointed out subsequently by others this last sentence should have read

The intervals (-∞, a] and [a, +∞) are generally regarded as closed (as Binmore says) but written as if they were open at the infinity end.
 
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Tesla23

Joined May 10, 2009
542
Good question, you were quick to spot the discrepancy.

Note that Binmore avoids writing [-∞ anywhere. He only puts open brackets (-∞ around infinities.
I did say that people get into difficulties when employing ∞. I have not seen any author use the closed bracket around infinities.

The difficulty is that the interval (-∞, +∞) or ]-∞, +∞[ is regarded as both open and closed, but never written [-∞, +∞]

The intervals (-∞, a) and (a, +∞) are generally regarded as closed (as Binmore says) but written as if they were open at the infinity end.
(-∞, a) and (a, +∞) are open and not closed
(-∞, a] and [a, +∞) are closed and not open

whereas
(-∞, +∞) is both open and closed as you point out.

There is no need for a 'generally regarded' comment, the definitions are clear, if the set has all it's accumulation points then it is closed, whereas if every point has a neighbourhood in the set then it is open.
 

Thread Starter

MichealY

Joined Apr 9, 2009
49
I agree with Tesla23 and Papabravo.

The intervals (-∞, a) and (a, +∞) are generally regarded as closed (as Binmore says) but written as if they were open at the infinity end.
The invtervals (-∞,a) and (a,+∞) are open and not closed,they have not got their all accumulation/limit point,say a.
 

studiot

Joined Nov 9, 2007
4,998
Yes it was pretty late here and I was trying to cut typing corners by copy/pasting.

I have added a correction to the post.

Sorry.
 
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