$\mathbb{R}$ carries both an unbounded and a bounded metric inducing the same topology

Refuted claim: boundedness of a metric space is determined by its topology (FALSE: boundedness of a metric space is determined by its topology).

The witness is the real line carrying two metrics at once:

$d(x,y) = |x - y| \qquad \text{and} \qquad \rho(x,y) = \min\{\, |x-y|,\ 1 \,\}.$

They induce the same topology, $(\mathbb{R}, d)$ is unbounded and has no diameter, and $(\mathbb{R},\rho)$ is bounded with diameter exactly $1$ ($\min(|x-y|, 1)$ on $\mathbb{R}$ has the usual topology and diameter at most $1$, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

• Which level of the hierarchy does preserve boundedness. Lipschitz equivalence does, since $\rho \le \beta d$ turns a $d$-ball into a $\rho$-ball of radius $\beta$ times as large (Topologically, uniformly and Lipschitz equivalent metrics on a set). The present pair is uniformly but not Lipschitz equivalent (On $\mathbb{R}$ the metrics $|x-y|$ and $\min(|x-y|,1)$ are uniformly but not Lipschitz equivalent), which is exactly the room in which boundedness is lost.
• Nor is the diameter a topological invariant, even up to a constant: on $\mathbb{R}$ the same construction with $\min\{d, c\}$ in place of $\min\{d, 1\}$, for any real $c > 0$, gives diameter $c$ while leaving the topology alone.
• The general statement behind this witness is that every metric space is boundedly remetrisable ($\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology), so the failure is not a peculiarity of the real line but the normal state of affairs.