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Results tagged with metric-spaces
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user 5903
A metric space is a pair $(X,d)$, where $X$ is a set and $d:X \times X \to \mathbb{R}$ satisfies the following conditions for all $x,y,z \in X$. (Symmetry) $d(x,y)=d(y,x)$. (Identity of Indiscernibles) $d(x,y)=0$ if and only if $x=y$. (Triangle Inequality) $d(x,y)+d(y,z) \geq d(x,z)$.
2
votes
Accepted
Can a Polish space have two different topologies?
Every Polish space is countable or of cardinality $\mathfrak{c}$.
Consequently, every countable Polish space has a homeomorphic copy with underlying set the natural numbers, and every uncountable one …
- 11.4k
2
votes
Accepted
A property on some unbounded metric spaces
Unboundedness guarantees that there is one sequence $(z_n)_n$ such that $d(x,z_n)\to\infty$ for all $x$.
That sequence also satisfies the second requirement via the triangle inequality:
$$
\frac{d(x,z …
- 11.4k
10
votes
Accepted
Is there a metric compactification that doesn't create new paths?
Here's a counterexample.
Let $B$ be a Bernstein set in the plane, so $B$ and its complement intersect every uncountable closed subset of $\mathbb{R}^2$.
Let $X$ be a metric compactification of $B$, wi …
- 11.4k
2
votes
End point compactification for metric spaces
Another possibility is to use proximities - or equivalently (totally bounded) uniformities: in the metric case one defines $A$ and $B$ to be 'close' (usually denoted $A\mathrel\delta B$) if $d(A,B)=0$ …
- 11.4k
2
votes
Extending homeomorphisms between compact metric subsets
As you can see from the comments the answer is: hardly ever.
As mentioned above the case of one-point sets necessitates the space being homogeneous. But that is not enough, say in $\mathbb{R}$ when y …
- 11.4k
2
votes
Accepted
Is every subgroup closed in this complete, nondiscrete topological group?
The metric induces the product topology, so the group $G$ is compact. The direct sum of $\mathbb{Z}$ many copies of $G'$ is a countable dense subgroup, but not the whole group, so it is not closed.
- 11.4k
16
votes
Accepted
Partition of unity without AC
The proofs rely, in the background, on Urysohn's Lemma, which follows from the Principle of Dependent Choices but is not provable without some Choice. It is false in the ordered Mostowski model, see
G …
- 11.4k