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 when related to time. 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)