A uniformly continuous map sends Cauchy sequences to Cauchy sequences

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric), let $f : X \to Y$ be uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point) and let $(x_k)$ be a Cauchy sequence in $X$ (Cauchy sequence in a metric space). Then $\big(f(x_k)\big)$ is a Cauchy sequence in $Y$.

Continuity alone does not suffice, and the failure is not marginal: a continuous map can send a Cauchy sequence to an unbounded one. The witness is named in the remarks below.

• This is the exact point where uniform continuity is indispensable. The $\delta$ produced in step 1.1 is chosen before any index is known, and it is then fed to the Cauchy condition. With ordinary continuity the $\delta$ would depend on a base point, and there is no base point available: the sequence has no limit in $X$ to serve as one. That is not a defect of the proof but the reason the statement is false for continuous maps; the witness is $x \mapsto 1/x$ on $(0,1)$ ($x \mapsto 1/x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1/(k+2))_{k \ge 0}$ to an unbounded one ↗).
• No completeness and no surjectivity is assumed, and $f$ need not be injective. The theorem is a statement about a single map and a single sequence.
• Consequences used later on this page. Uniformly equivalent metrics have the same Cauchy sequences (Topologically, uniformly and Lipschitz equivalent metrics on a set), because uniform equivalence says exactly that both identity maps are uniformly continuous; and a uniformly continuous map defined on a dense subspace transports the Cauchy sequences that approximate a point, which is what makes A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space possible.