This might seem like a silly question, but is there a number that's right in the middle of this interval $(0, 1)$?

And the half-open intervals: $(0, 1]$, $[0, 1)$? I know for a fully closed interval $[0, 1]$ it's $1/2$ because that's half the length of the interval but for $(0 ,1)$, I can't say. What's the interval's length? Certainly not $1$.

Edit: In response to the comment below: I said, in my mind, it can't be $1$ because that that length of the interval $[0, 1]$, and that's not the same as $(0, 1)$. I suppose it approaches 1, but that's not the same, or is it?

And I think by 'length' I mean the the difference in the end points: $|x_2 - x_1|$. $(0, 1)$ does't include $0$ or $1$. There are no end-points to subtract!

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    $\begingroup$ What definition of length are you using? Most standard definitions of length would say $(0,1)$ has length one. Similarly, what do you mean by "in the middle"? $\endgroup$ – Ryan Budney Dec 1 '12 at 21:30
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    $\begingroup$ Yes, the answer is $1/2$ (according to the standard definition of length). The length of all intervals $(0,1)$, $[0,1]$, $(0,1]$ and $[0,1)$ is $1$. $\endgroup$ – Yury Dec 1 '12 at 21:33

To answer this question, you need to first answer the following question. $$\text{What is the `length' of a single point?}$$ If you take it to be zero, then note that $$[0,1] = \{0\} \cup (0,1) \cup \{1\}$$ And assuming that length of these disjoint sets can be added, we get that the length of the interval $(0,1)$ is the same as the length of the interval $[0,1]$.

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    $\begingroup$ I don't quite follow your post. Can't the entire interval [0, 1] be partitioned into infinitely-many singleton sets, with each adding length zero and thus $[0, 1] = \{0\} \cup \dots \cup \{1\} = 0$? What am I missing? $\endgroup$ – Mark Dec 1 '12 at 21:58
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    $\begingroup$ @Mark What you are asking is a very good question. The notion of length is formalized in a branch of analysis called measure theory according to which if $A_k$'s are countably disjoint sets, then it is indeed true that $$\text{Length}(A_1 \cup A_2 \cup \cdots \cup A_n \cup \cdots) = \text{Length}(A_1) + \text{Length}(A_2) + \cdots + \text{Length}(A_n) + \cdots$$ However, in your case $[0,1]$ contains uncountably many points. Hence, though it is true that $$[0,1] = \cup_{x \in [0,1]} \{x\}$$ it is incorrect to say that $\endgroup$ – user17762 Dec 1 '12 at 22:01
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    $\begingroup$ $$\text{Length} (\cup_{x \in [0,1]} \{x\}) = \sum_{x \in [0,1]} \text{Length}({x})$$ However, for instance, it is correct to say that the "length" of the set of natural numbers is $0$ since $$\text{Length}(\bigcup_{n=0,1}^{\infty} \{n\}) = \sum_{n=0}^{\infty} \text{Length}(\{n\}) = \sum_{n=0}^{\infty} 0 = 0$$ since the natural number set can be written as a countable union of disjoint single element sets, the length of whom are zero. $\endgroup$ – user17762 Dec 1 '12 at 22:04
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    $\begingroup$ OK, so because my partitioning isn't countable, what I did wasn't right. $\endgroup$ – Mark Dec 1 '12 at 22:08
  • $\begingroup$ @Mark Yes. precisely. $\endgroup$ – user17762 Dec 1 '12 at 22:09

See Wikipedia for a discussion of an interval. The open interval $(1, 0)$ is a bounded set,

and therefore has diameter (length) = $\sup((0, 1)) - \inf((0, 1)) = |1 - 0| = 1$.

Also note that the intervals $\left(0, 1\right]$ and $\left[0, 1\right)$ are also bounded and each is of length $1$.

Half the length of any interval of length $1$ is $\frac12$.

So $\frac12$ could be said to be in the "middle" of each of the above intervals,

since $\left(0, \frac12\right)$, $\left(0, \frac12\right]$, $\left[0, \frac12\right)$, $\left[0, \frac12\right]$, $\left(\frac12, 1\right)$, $\left(\frac12, 1\right]$, $\left[\frac12, 1\right)$, and $\left[\frac12, 1\right]$

each have diameter = length = $\frac12$.


One sensible definition for the middle point of a bounded set $A \subseteq \mathbb{R}$ could be

$$\operatorname{middle}(A) = \frac{\sup(A) + \inf(A)}{2}$$

This is a good definition for intervals, and the middle point of $(0,1)$ would be $1/2$ with this definition. However, when $A$ is not an interval, $\operatorname{middle}(A)$ might not be a point of $A$.


Any interval on the real line can be written as $(a,b)$,$[a,b)$,$(a,b]$ or $[a,b]$ with $a\leq b$ being extended real numbers (the set of all real numbers with $+\infty$ and $-\infty)$ called its endpoints. The length of the interval (possibly $0$ or $+\infty$) can then be defined as the difference between its endpoints. If we use Lebesgue measure on the real line, it can be shown that the measure of an interval is simply its length.

If number in the middle of an interval means the length on either side is equal, then indeed $\frac{1}{2}$ would be in the middle of the intervals $(0,1)$,$[0,1)$,$(0,1]$ and $[0,1]$, because for example the length of $(0,\frac{1}{2})$ and $(\frac{1}{2},1)$ are each $\frac{1}{2}$.

  • $\begingroup$ You have to be even more careful: what exactly do you mean by "endpoints"? $\endgroup$ – Adam Rubinson Dec 1 '12 at 21:36
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    $\begingroup$ @Adam: Obviously, the endpoints of all the non-empty intervals $[a,b]$, $[a,b)$, $(a,b]$, and $(a,b)$ are $a$ and $b$. If you can't suggest any other remotely plausible interpretation, then you have to accept this! $\endgroup$ – TonyK Dec 1 '12 at 21:47
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    $\begingroup$ OP's question is pretty simple so he might not be used to there being a difference between inf and min, and using the term "endpoints" might confuse him because "endpoint" seems to imply 1 is in (0,1). $\endgroup$ – Adam Rubinson Dec 1 '12 at 21:50

The "length" of this interval can't be less than 1 (see below), and it can't be greater than 1 (it's contained in an interval of length 1), so the length must be 1.

Suppose the length is some number $1- \varepsilon $ where $\varepsilon > 0$. Then we can plop an interval of this length right in the middle of our (0,1) interval, so we have $[\frac{\varepsilon}{2},1- \frac{\varepsilon}{2}] \subset (0,1)$. BUT, we're still leaving out infinitely many points (namely, the intervals $(0,\frac{\varepsilon}{2})$ and $(1-\frac{\varepsilon}{2},1)$. Thus, saying its length is anything less than 1 wouldn't make any sense, so we can say that the length of the interval must be at least 1.

Coupled with the fact that the interval is of size no greater than 1, we have $$1 \leq size(0,1) \leq 1$$ so that the size is exactly 1.

Now, to answer the original question, this would make the midpoint of the interval exactly $\frac{1}{2}$.

The word you're looking for when you say "length" of the interval, I believe, is called the diameter of the set.


No one has actually defined the length of an interval precisely. Any interval can be written in exactly on of the forms: (a,b) or (a,b] or [a,b) or [a,b], where b >= a. In any of the 4 cases, call the interval I, a subset of R. Then define the length of the interval to be sup{x: x is in I} - inf{x: x is in I}.

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    $\begingroup$ I suppose you mean no one in this thread has defined the length of an interval precisely? Even if you do mean that, it's not true (and wasn't true when you wrote it -- see for instance Will Hunting's answer). $\endgroup$ – TonyK Dec 1 '12 at 21:43

protected by Qiaochu Yuan Dec 2 '12 at 1:53

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