If $X$ is a metric space such that any metric space $Y$ , which is a homeomorphic image of $X$ , is complete , then is $X$ compact?

Question

Let $X$ be a compact metric space , then it is easy to show that every homeomorphic image metric space of $X$ is complete . Is the reverse true ? That is if $X$ is a metric space such that any metric space $Y$ , which is a homeomorphic image of $X$ , is complete , then is $X$ compact ? I know that $(X,d)$ is homeomorphic with the bounded metric space $(X , \dfrac{d}{1+d})$ and also $(X , \min \{1,d\})$ , but I cannot get a totally bounded homeomorphic image . Please help .

Answer 1

Partial answer.

If $X$ is a separable metric space, then it embeds in $Z=[0,1]^{\omega}$ ($\omega$ is the cardinal of $\mathbb{N}$) by the Urysohn metrization theorem (Kelley, theorem 4.16). If $X$ is not compact, then the image $Y$ of the embedding is not closed in $Z$ (because $Z$ is compact). Note, however, that $Z$ is metrizable, so $Y$ is not complete in the induced metric, or it would be closed.

Hence a non compact separable metric space admits an equivalent metric in which it is not complete.