The following question came up in my 245A class today:

Is it possible to express a non-closed interval in the real line, such as [0,1), as a countable union of disjoint closed intervals?

I was not able to answer the question immediately, but by the end of the class some of the students had come up with an answer.  It is actually a nice little test of one’s basic knowledge of real analysis, so I am posing it here as well for anyone else who is interested.  Below the fold is the answer to the question (whited out; one has to highlight the text in order to read it).

The answer to the question is… no.

First of all, it suffices to prove the claim for an open interval such as (0,1).  For, if one partitions a half-open interval such as [0,1) into closed intervals, then after selecting any one of these intervals [a,b], the open interval (b,1) must then also be partitioned into countably many disjoint closed intervals.

By mapping (0,1) homeomorphically to the real line, it suffices to show that the real line R cannot be partitioned into closed intervals.

As each interval is bounded, we need an infinite number of intervals I_n = [a_n,b_n] to cover the real line.  Now consider the set

E := \bigcup_{n=1}^\infty \{a_n,b_n\}

consisting of the endpoints of the intervals I_n.  Clearly, E is countably infinite.  Also, as the I_n form a disjoint cover of E, E is the complement of the open set \bigcup_{n=1}^\infty (a_n,b_n) and is hence closed.  Finally, we claim that E is perfect: that not only is E closed, but every point in E is a limit point in E.  Indeed, if x lies in E, then x is either the left or right endpoint of an interval I_n, but it is not both.  If it is, say, the right-endpoint of an interval, then by approaching x from the right we see that x is the limit of the left endpoint of intervals to the right of x.

Now we appeal to a general theorem (a special case of the even more general Baire category theorem) that asserts that a perfect subset of a complete metric space cannot be countably infinite.  A proof is as follows.  Suppose for contradiction that E = \{ x_1, x_2, x_3, \ldots\} is a countably infinite perfect set.  Let B_1 be any closed ball of positive radius whose centre lies in E (e.g. one can take the closed ball of radius 1 centered at x_1).  Using the fact that x_1 is a limit point, one can then find a closed ball B_2 inside B_1 whose centre still lies in E, but which is disjoint from x_1, and whose radius is at most half that of B_1.   (Indeed, if x_1 does not already lie in the interior of B_1, one simply has to shrink B_1 by a little bit and recenter to another element of E to move x_1 off the boundary of B_1; otherwise, one uses the limit point property to find another point in the interior of B_1 other than x_1, and makes a small closed ball around that.)  Then one can find a closed ball B_3 inside B_2 of radius at most half that of B_2 whose centre still lies in  E, but is now also disjoint from x_2 in addition to x_1.  Continuing in this fashion we find a sequence B_1, B_2, B_3, \ldots of nested closed balls with radii going to zero, each of which has centre in E.  The centres of these balls then form a Cauchy sequence, which by completeness, must converge to a limit in E that also lies in \bigcap_{n=1}^\infty B_n.  But by construction, \bigcap_{n=1}^\infty B_n lies outside each element x_n of E, and is thus disjoint from E, a contradiction.

Several of the students also came up with a proof based on the fact that the set E had the topological structure of a Cantor set, which is necessarily uncountable.  This is close to the proof given above, though the above proof has the advantage that it is more readily generalisable to higher dimensions, showing for instance that a non-closed box cannot be expressed as the countable union of closed boxes.  (For the generalisation, one has to use the full strength of the Baire category theorem, rather than just relying on the theory of perfect sets.)  UPDATE: as pointed out in comments, the higher-dimensional result can also be easily deduced from the one-dimensional result by looking at a slice.

Note that if one replaces the reals in the question by the rationals, then the answer to the question becomes “yes” (this is easiest to see if one uses closed intervals of zero length, though one can also concoct examples using positive length intervals also).  This shows that any proof must use a property of the real line, such as completeness, that is not shared by the rationals, and so it is unlikely that there is any elementary proof that is significantly shorter than the one given above.