# What is NOT an interval?

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• Oct 29th 2009, 10:26 AM
Furbylicious
What is NOT an interval?
Intervals are [a,b], right?

What is NOT an interval? Is it a single point?
• Oct 29th 2009, 10:31 AM
Turiski
Lots of things aren't intervals.

I'm not sure about points, but if you could give a little context I think that would help.
• Oct 29th 2009, 10:47 AM
Furbylicious
Quote:

Originally Posted by Turiski
Lots of things aren't intervals.

I'm not sure about points, but if you could give a little context I think that would help.

Well like an interval [0,1] is an interval, as is (-2, 100000], which u can draw on a number line.

What is not classed as an interval?
• Oct 29th 2009, 10:53 AM
Turiski
An interval is a set of real numbers that lie between two given numbers.

so [a,a] is a valid interval. So is [a,a), it's just that it's an empty one.

• Oct 29th 2009, 11:00 AM
Furbylicious
Give an example of something that isn't an interval then.
• Oct 29th 2009, 11:03 AM
Turiski
A formal proof is not an interval.

[Unless you can find some crazy way to reduce it down to a set of 'continuous' numbers (probably not the right term), which, by the way, would be totally awesome and I'd love to see that.]
• Oct 29th 2009, 11:25 AM
Swlabr
Quote:

Originally Posted by Furbylicious
Give an example of something that isn't an interval then.

$(1,2) \cup (3,4)$ is not, and neither is $[1,3] \setminus \{2\}$.

An interval is a subset of $\mathbb{R}$ where if you take two numbers from the subset every number between them is in the subset: $I \subseteq \mathbb{R}, \text{ } a, b \in I$, then $c \in I \text{ } \forall \text{ } a \leq c \leq b \in \mathbb{R}$. So as soon as you take one or more of these numbers out you stop having an interval.
• Oct 29th 2009, 12:06 PM
Plato
Quote:

Originally Posted by Furbylicious
Intervals are [a,b], right?
What is NOT an interval? Is it a single point?

Some authors define an interval in $R^1$ to be a non-degenerate( non-singleton) connected set.
So basically there are eight types, four bounded and four unbounded:
$(a,b),~(a,b],~[a,b),~[a,b],~(- \infty ,b), ~(- \infty ,b],~(a, \infty),~\& ~[a, \infty)$

Because it is required that an interval contain at least two points, $\{a\}$ is not an interval.
• Oct 30th 2009, 07:03 AM
HallsofIvy
unions of non-overlapping intervals are not intervals:
$[0, 1]\cup [2, 3]$ is not an interval.

The set of all rational numbers in [0,1] not only is not an interval but contains no interval because, between any two rational numbers, there is an irrational number.
• Oct 30th 2009, 10:23 AM
Failure
Quote:

Originally Posted by HallsofIvy
unions of non-overlapping intervals are not intervals:
$[0, 1]\cup [2, 3]$ is not an interval.

The set of all rational numbers in [0,1] not only is not an interval but contains no interval because, between any two rational numbers, there is an irrational number.

I certainly agree that the set of all rational numbers in [0,1] is not a real interval. But as a subset of the rational numbers it actually is an interval: it is a rational interval.
• Oct 30th 2009, 10:35 AM
Failure
Quote:

Originally Posted by Plato
Some authors define an interval in $R^1$ to be a non-degenerate( non-singleton) connected set.

But that's the definition used only by some authors, as you write. Intervals are not at all generally defined that way. It is, in my humble opinion (and I could quote several books in my side), much more natural to define intervals as the connected subsets of $\mathbb{R}$ - without slapping any further condition(s) on top of that basic requirement.

Quote:

So basically there are eight types, four bounded and four unbounded:
$(a,b),~(a,b],~[a,b),~[a,b],~(- \infty ,b), ~(- \infty ,b],~(a, \infty),~\& ~[a, \infty)$

Because it is required that an interval contain at least two points, $\{a\}$ is not an interval.
But, as I wrote above, not a few mathematicians would look askance at you if you told them that $\{a\}$ is not an interval...
• Oct 30th 2009, 10:46 AM
Plato
Quote:

Originally Posted by Failure
But that's the definition used only by some authors, as you write. Intervals are not at all generally defined that way. It is, in my humble opinion (and I could quote several books in my side), much more natural to define intervals as the connected subsets of $\mathbb{R}$ - without slapping any further condition(s) on top of that basic requirement.

But, as I wrote above, not a few mathematicians would look askance at you if you tell them that $\{a\}$ is not an interval...

Please give me some names. I can not think any offhand.
• Oct 30th 2009, 11:07 AM
Failure
Quote:

Originally Posted by Plato
Please give me some names. I can not think any offhand.

Well, Walter Rudin does it repeatedly, in "Principles of Mathematical Analysis" as well as in "Real & Complex Analysis". Jean Dieudonné does it, in "Foundations of Modern Analysis". And if Dieudonné does it I take it for granted that all the other members of the Bourbaki Group would at least go along with that. (I think it would be not particularly useful to you if I mentioned German authors: but being Swiss German, many of my books are in .. you guessed it .. German.)

Wikipedia also defines intervals in such a way that the "interval" [a,a] is just taken to be the singleton set {a}.

Defining intervals this way ("my way", i.e. Rudin's and Dieudonné's way), allows one to say something simple in a simple way: the image of an interval under a continuous function $f:\, \mathbb{R}\rightarrow \mathbb{R}$ is an interval. You could not say the exact same thing as succinctly if you defined interval in such a way as to exclude singleton sets.

Of course, all definitions are arbitrary - but some definitions are more arbitrary than others....
• Oct 30th 2009, 12:23 PM
Plato
Quote:

Originally Posted by Failure
Well, Walter Rudin does it repeatedly, in "Principles of Mathematical Analysis" as well as in "Real & Complex Analysis". Jean Dieudonné does it, in "Foundations of Modern Analysis".

How very odd. Your edditions/translations of Rudin and Dieudonne differ so from mine.
In both 'Big Rudin' and 'Little Rudin' that I have, this definition appears:
“The segment (a,b) is the set of all real numbers, x , such that a<x<b.”
Then he notes that a interval is a segment that includes its endpoints.
To me that rules out {a} being an interval.

Virtually the same statement appears on page 17 of my translation of Dieudonne.

This is odd, don't you think?
• Oct 30th 2009, 04:21 PM
tonio
Quote:

Originally Posted by Plato
How very odd. Your edditions/translations of Rudin and Dieudonne differ so from mine.
In both 'Big Rudin' and 'Little Rudin' that I have, this definition appears:
“The segment (a,b) is the set of all real numbers, x , such that a<x<b.”
Then he notes that a interval is a segment that includes its endpoints.
To me that rules out {a} being an interval.

Perhaps, but I don't think it does: even accepting the above version, which btw I cannot find neither in Rudin's "Real and Complex Analysis" nor in "Principles of Mathematical Analysis", the singleton {a} would be the interval (a,a) with its endpoints included

Virtually the same statement appears on page 17 of my translation of Dieudonne.

In Dieudonne's "Foundations of Modern Analysis", enlarged and corrected edition 1969, page 17, I read the definition of a closed interval, and then he even adds "...(for a=b the notation [a,a] means the one-point set {a})". If at all I could conclude that Dieudonne's accepting the singletons as closed intervals, but perhaps someone else would understand this otherwise.

This is odd, don't you think?

Interpretations sometimes, but nowhere coud I find any source requiring any minimal number of points for something to be considered a real interval, and I think the huge majority of mathematicians would agree that {a} = [a,a] is a closed interval, in particular since the usual topology on $\mathbb{R}$ is Hausdorff and thus every singleton is a closed set, but of course: I could be wrong.

Tonio
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