Many people think that suprema and infima are ghost like entities which do not really appear under "real" circumstances and are just Mathematical curiosities. The infimum of 0 minutes and the supremum of 10 minutes on the proof below, are very much tangible and give one a very clear idea of how difficult they are to be achieved. They are, in fact, non-attainable. This however should not scare us away from such definitions, as they have very "real" meanings after all.
For regular subsets of R1, we use the standard notations sup E and inf E for the supremum (least upper bound) and infimum (greatest lower bound) of E. In case sup E belongs to E, it will be called max E; similarly if inf E belongs to E it will be called min E. {ak} reads a sub k. If {ak}[k=1,∞] is a sequence of points in R1, let bj=sup(k≥j)(ak) and cj=inf(k≥j)(ak), j=1,2,3,...Then -∞≤cj≤bj≤∞ and {bj} and {cj} are monotone decreasing and increasing respectively; that is, bj≥bj+1 and cj≤cj+1. Define limsup(k®∞) and liminf(k®∞) as follows:
limsup(k®∞) ak=lim(j®∞)
bj=lim(j®∞) {sup(k≥j) ak},
liminf(k®∞) ak=lim(j®∞)
cj=lim(j®∞) {inf(k≥j) ak}.
Consider the set T={total time taken in a 5 (each player) minute speed chess game, at the minute of flag fall, excluding indeterminate draws (either by repetition of moves or insufficiency of forces.)}
Prove that liminf T=0 and that limsup T =10. Use the following:
(a) L=limsup(k®∞) ak if and only if (i) there
is a subsequence {akj} of {ak} which converges to
L, and (ii) if L&;gt;L, there is an integer K such that ak<L'
for k≥K. (for suprema) and
(b) M=liminf(k®∞) ak if and only if (i) there
is a subsequence {akj} of {ak} which converges to
M, and (ii) if M'<M, there is an integer K such that ak>M'
for k≥K. (for infima)
Proof for the limsup. Let G be a game with time t in T. We use (a).
(i) Consider the sequence of games {gk} which are identical
to G, except that each move is played in time 10/(#of moves)-(time of move
in G)/k, where k=1,2,3,... Clearly this sequence of games gives rise to
a sequence of times {tk}. The sequence {tk} has members
in T which is a subset of all possible time points {ak}=T. So
clearly {tk} is a subsequence of {ak}. It clearly
converges to L=10. [The total time of those games will be (#of Moves)*10/(#of
moves)-(time of move 1 in G)/k-(time of move 2 in G)/k-...)=10-(total time
of G)/k®10].
(ii) Let L>10. Then one of the flags will fall first. This flag will
show 5 exactly. The other flag will necessarily show x < 5, since it
hasn't fallen yet. So {ak}<5+x<10<L' for all k (The
flags CANNOT fall simultaneously because it is always ONE player's turn).
QED.
You now prove the dual for the liminf. Extra credit: Are 0 and 10 actually IN T?? (i.e. are they THE minimum and maximum respectively?? Justify your answer)