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 R, 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. {a_k} reads a sub k. If {a_k}_[k=1,∞] is a sequence of points in R1, let b_j=sup_(k≥j)(a_k) and c_j=inf_(k≥j)(a_k), j=1,2,3,...Then -∞≤c_j≤b_j≤∞ and {b_j} and {c_j} are monotone decreasing and increasing respectively; that is, b_j≥b_j+1 and c_j≤c_j+1. Define limsup_(k->∞) and liminf_(k->∞) as follows:

1. limsup_(k->∞) a_k=lim_(j->∞) b_j=lim_(j->∞) {sup_(k≥j) a_k}.
2. liminf_(k->∞) a_k=lim_(j->∞) c_j=lim_(j->∞) {inf_(k≥j) a_k}.

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:

1. L=limsup_(k->∞) a_k if and only if (i) there is a subsequence {a_kj} of {a_k} which converges to L, and (ii) if L>L, there is an integer K such that a_k<L' for k≥K. (for suprema)
2. M=liminf_(k->∞) a_k if and only if (i) there is a subsequence {a_kj} of {a_k} which converges to M, and (ii) if M'<M, there is an integer K such that a_k>M' for k≥K. (for infima)

Proof:

For the limsup: Let G be a game with time t in T. We use 1).

1. Consider the sequence of games {g_k} 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 {t_k}. The sequence {t_k} has members in T which is a subset of all possible time points {a_k}=T. So clearly {t_k} is a subsequence of {a_k}. It clearly converges to L=10, because 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.
2. 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 (the flags cannot fall simultaneously because it is always one player's turn), since it hasn't fallen yet. So {a_k}<5+x<10<L' for all k .

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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