Suprema and Infima in real life

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>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)

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