Prove equivalence of definitions of "dense" Prove that these two statements are materially equivalent (that is, one statement can be derived or proven from the other). Read below the statements for background information if you need it.


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*Given a metric space $M$, $M$ is dense if and only if for every point $x\in M$, and for every positive real number $\delta$, there exists a point $t\in M$ (with $t\ne x$) such that $t$ is in the delta-neighborhood $V_\delta(x)$ surrounding $x$.

*Given a metric space $M$, $M$ is dense if and only if for every two points $x,y \in M$ and $x\ne y$, there exists another point $t$ that is between $x$ and $y$, with $x\ne t\ne y$. Between is defined rigorously as $d(x,t)+d(t,y) = d(x,y)$.


Background Information
Metric Spaces
A metric space $M$ is a set paired with a function $d\colon M^2 \mapsto \mathbb{R}$ (thought of as “distance”) defined such that for all elements (“points”) $a,b,c \in M$,


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*$d(a,a)=0$ (the distance between any point and itself is $0$)

*$d(a,b)=d(b,a)$ (the distance between any two points is the same in both directions)

*$d(a,b)+d(b,c)\ge d(a,c)$ (Triangle Inequality)


Furthermore, a fourth property can be derived from the first three: that $d(a,b)\ge 0$. tl;dp.
Delta-Neighborhoods
Given a metric space $M$, a point $c\in M$, and a distance $\delta\in\mathbb{R}$, the delta-neighborhood surrounding $c$ $V_\delta(c)$ is the subset of all $s\in M$ such that $d(c,s)<\delta$. In other words, it's the set of all points closer than $\delta$ units away from $c$. On $\mathbb{R}$, this is visualized as the interval $(c-\delta,c+\delta)$; on $\mathbb{R}^2$, it's visualized as a disc; on $\mathbb{R}^3$, as a ball.
 A: What you call a metric is usually known as a metric space, but this is a minor point. 


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*Assuming that in the first property we also require $t\ne x$ (to make it nontrivial), the property amounts to saying that $M$ has no isolated points.

*The second definition describes a convex metric space.


Neither implies the other. On the real line, $[0,1]\cup [2,3]$ has no isolated points, but is not convex. On the other hand, $\{0\}\cup (1,2)$ has an isolated point but is convex (according to the above definition). 
If you also required $M$ to be complete in the second definition, then it would imply the first one. Sketch of proof: given $a,b\in M$, let $d=d(a,b)$ and consider the set
$$T= \{t\in (0,1): \exists c \text{ such that } d(a,c)=td, \ d(b,c)=(1-t)d\}$$
If $\inf T>0$, then, using completeness, we can show that $\inf T\in T$, that is there exists $c$ as above with $t=\inf T$. But then convexity is violated between $a$ and $c$. Thus, $\inf T=0$, which implies that $a$ is not isolated.
