sentences:  It is almost trivial to prove that the product of two sequentially compact spaces is sequentially compact—one passes to a subsequence for the first component and then a subsubsequence for the second component.An only slightly more elaborate "diagonalization" argument establishes the sequential compactness of a countable product of sequentially compact spaces.However, the product of continuum many copies of the closed unit interval (with its usual topology) fails to be sequentially compact with respect to the product topology, even though it is compact by Tychonoff's theorem (e.g., see ).