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. {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:

limsup_{(k®∞)} a_{k}=lim_{(j®∞)}
b_{j}=lim_{(j®∞)} {sup_{(k≥j)} a_{k}},

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:

(a) 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) and

(b) 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 (a).

(i) 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. [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 {a_{k}}<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)