open set in contains an open rectangle
Theorem Suppose is
equipped with the usual topology induced by the open balls of the
Euclidean metric.
Then, if is a non-empty open set in , there
exist real numbers for such that
and is a subset of .
Proof.
Since is non-empty, there exists some point
in . Further, since is a topological space, is contained in
some open set. Since the topology has a basis consisting of
open balls, there exists a and such that
is contained in the open ball .
Let us now set and
for all .
Then can be
parametrized as
For an arbitrary point in , we have
so , and the claim follows.
Title | open set in contains an open rectangle![]() |
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Canonical name | OpenSetInmathbbRnContainsAnOpenRectangle |
Date of creation | 2013-03-22 14:07:46 |
Last modified on | 2013-03-22 14:07:46 |
Owner | matte (1858) |
Last modified by | matte (1858) |
Numerical id | 5 |
Author | matte (1858) |
Entry type | Theorem |
Classification | msc 54E35 |
Related topic | Interval![]() |