from the original sequence to a subsequence times—which is still a subsequence of the original sequence—on which each coordinate sequence converges, hence the subsequence itself is convergent. Alternative proof There is also an alternative proof of the Bolzano–Weierstrass theorem using nested intervals. We start with a bounded sequence : Because we halve the length of an interval at each step, the limit of the interval's length is zero. Also, by the nested intervals theorem, which states that if each is a closed and bounded interval, say with then under the assumption of nesting, the intersection of the is not empty. Thus there is a number that is in each interval . Now we show, that is an accumulation point of . Take a neighbourhood of . Because the length of the intervals converges to zero, there is an interval that is a subset of . Because contains by construction infinitely many members of and , also contains infinitely many members of . This proves that is an accumulation point of . Thus, there is a subsequence of that converges to . Sequential compactness in Euclidean spaces Definition: A set is sequentially compact if every sequence in has a convergent subsequence converging to an element of . Theorem: is sequentially compact if and only if is closed and bounded. Proof: (sequential compactness implies closed and bounded) Suppose is a subset of with the property that every sequence in has a subsequence converging to an element of . Then must be bounded, since otherwise the following unbounded sequence can be constructed. For every , define to be any arbitrary point such that . Then, every subsequence of is unbounded and therefore not convergent. Moreover, must be closed, since any limit point of , which has a sequence of points in converging to itself, must also lie in . Proof: (closed and bounded implies sequential compactness) Since is bounded, any sequence is also bounded. From the Bolzano-Weierstrass theorem, contains a subsequence converging to some point . Since is a limit point of and is a closed set, must be an element of . Thus the subsets of for which every sequence in A has a subsequence converging to an element of – i.e., the subsets that are sequentially compact in the subspace topology – are precisely the closed and bounded subsets. This form of the theorem makes especially clear the analogy to the