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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) $d(x,y)+d(y,z) \geq d(x,z)$.
1
vote
radius-diameter relationship of balls in metric spaces
There is a way of enlarging a metric space $(X,d)$ to a larger metric space $(Y,d)$ so that every ball $B_r(x)$ has diameter $2r$ if and only if there are points $y,z\in Y$ where $d(x,y)=d(x,z)=r,d(y, …
8
votes
Accepted
Locally compact space that is not topologically complete
Every locally compact metric space can be given a compatible complete metric.
Suppose that $X$ is a locally compact metric space. Then $X$ is paracompact, so $X$ is a disjoint union of $\sigma$-compac …
3
votes
A property for maps between metric spaces
Under mild conditions on the metric space $(X,d)$, we can assure that either $f$ is constant or $f$ is injective, so $f$ is either trivial or behaves similar to an isometry. Let $\simeq$ be the equiva …
5
votes
Baire Category Theorem for complete uniform spaces
It is very common for a topological space to be a complete uniform space in some uniformity, but it is less common for a topological space to satisfy the Baire category theorem since the proof of the …
5
votes
Uniform density of Lipschitz maps is space of continuous function — for general metric spaces
I claim that $X,Y$ are metric spaces with $X$ compact, and if $X$ or $Y$ is zero-dimensional, then every continuous function $f:X\rightarrow Y$ can be uniformly approximated by a locally constant func …
14
votes
Accepted
Uniform density of Lipschitz maps is space of continuous function — for general metric spaces
Let $(X,\rho)$ be a compact metric space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ ca …
3
votes
Accepted
Definition of $F_{\sigma}$ sets in terms of $\varepsilon$?
Suppose that $(X,d)$ is a complete metric space. Then a subset $G\subseteq X$ is a $G_{\delta}$-set precisely when $G$ can be given a complete metric which induces the subspace topology on $G$. The no …