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In the following, the terms of openness and closure of sets are destructed.

A disk without its boundary conventionally represents an open set, because then each point of it has a neighbourhood that lies completely in this set. Here, the idea is based on that, if the points are considered on a half-line, starting at the centre of the disk, it must always be considered a real neighbourhood for each point on this half-line towards the boundary.

In fact, however, "the end of the flagpole" must sometime be reached. So there must be a point in the interior of the disk, which has no neighbourhood in this interior. Therefore, the term openness for sets is to criticise. If the unit disk is considered around the origin of ordinates, so the last point of the half-line [0, 1[, dually represented, is the point 0,1 and the next point is the boundary point 1.

Between these two points lies no other point. The former has no neighbourhood that lies in the interior of the disk, though it is an interior point. For this reason, the disk without boundary is also closed, since the last points of the half-line form, beginning from the centre of the disk, just the closure as boundary. It can be easily shown from the exterior that the term closure for sets in Euclidean space is meaningless, since the neighbourhoods do not exist there for the boundary points of the latter.

Analogously it can be shown that at least in Euclidean space every open set is also closed, what reduces these terms to absurdity. This has not been further annoying, since infinitesimal quantities so far were not considered in a differentiated manner, thus in particular the numbers 1 and 0,1 were equated. However, this is incorrect as is explained in set theory, since otherwise algebraicity (1) and transcendence (0,1) are equated.

The absurd can also be illustrated by the fact that an infinite intersection of open sets such as all open concentric disks can form a closed set (the common centre of the disk) and an infinite union of closed sets again can build an open set as an open disk does, as a union of all of its points as closed sets.

A set consisting of a single point is therefore open, because every neighbourhood, since consistently being 0-dimensional and one-dimensional for intervals, also consists of one point. Therefore, the empty set is also closed, and as a consequence also the whole Euclidean space is closed. Using spheres, that what has been said can easily be generalised to higher dimensions. The terms of the inner and outer point remain meaningful however, if any infinitesimal radiuses are permitted.

Since an absurd and meaningless special case also makes the general case here absurd or meaningless, one also for metric and topological spaces openness or closure of sets should not be considered, particularly since the definition of a topological space appears oddly content-free and arbitrary, while the terms of interior and exterior point as well as boundary point are still useful and appropriate.

The neighboured boundary points of [0, 1] and ]0, 1[ (see nonstandardanalysis) especially have not the Hausdorff property. So not every metric space can be a Hausdorff space and the symmetric and (pre-) regular spaces are limited. Therefore, ℂ^{n} resp. ℝ^{n} with n ∈ ^{κ}ℕ* is a Kolmogorov space. The situation is only different in (imprecise) conventional mathematics.

© 2013 by Boris Haase

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