To prove a set is open, for any point x in a given set A defined by a given metric space, means that we need prove there exists a neighborhood N (or you might hear the terminology open ball, or open set, or other terms of a similar sort) around such point where N is contained in A...or another way of putting this, all points of N are in A, or N may be written as an open set such that N is contained in A.
Thus in one recently read proof, the customary manner of proving this would be starting with the assumption of a given open ball of radius r, and then having shown that from the metric of any point |a-x| < r means that a is an element of A. Here, proving that a set is open need not mean that every possible neighborhood (open set) is contained in A for a given point, but that at least one exists. One such technique that I've seen, for example, involved to defining a radius r by the given the elements of the neighborhood metric. Thus something like |a-x| < 1/|x^-1| and then from this proving that a were an element of A. In this case, set inclusion by the problem example, were implied by definition of a term of invertibility of a given element in the set, and invertibility were determined indirectly from continuity. A presiding assumption for the given set were that |u - e| < 1 where e where a unit element (i.e. where x and e in A means that x*e = e*x = x), meant by continuity that u where in invertible (an assumption here) and thus by definition of A (in this where A were a set of invertible elements), we could see from the defined neighborhood mentioned earlier that |a*x^-1 - e| < 1 would mean in turn that the product a*x^-1 is then invertible. Then using one other definition, namely, the product of invertible elements is in turn invertible should mean that (a*x^-1) * x being invertible means a is invertible.
Thus a bit of multi faceted logic here working from the presumption of a given defined neighborhood which in turn leads to an element being included in a given set which in turn proves a set being open.
Thus in one recently read proof, the customary manner of proving this would be starting with the assumption of a given open ball of radius r, and then having shown that from the metric of any point |a-x| < r means that a is an element of A. Here, proving that a set is open need not mean that every possible neighborhood (open set) is contained in A for a given point, but that at least one exists. One such technique that I've seen, for example, involved to defining a radius r by the given the elements of the neighborhood metric. Thus something like |a-x| < 1/|x^-1| and then from this proving that a were an element of A. In this case, set inclusion by the problem example, were implied by definition of a term of invertibility of a given element in the set, and invertibility were determined indirectly from continuity. A presiding assumption for the given set were that |u - e| < 1 where e where a unit element (i.e. where x and e in A means that x*e = e*x = x), meant by continuity that u where in invertible (an assumption here) and thus by definition of A (in this where A were a set of invertible elements), we could see from the defined neighborhood mentioned earlier that |a*x^-1 - e| < 1 would mean in turn that the product a*x^-1 is then invertible. Then using one other definition, namely, the product of invertible elements is in turn invertible should mean that (a*x^-1) * x being invertible means a is invertible.
Thus a bit of multi faceted logic here working from the presumption of a given defined neighborhood which in turn leads to an element being included in a given set which in turn proves a set being open.
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