# Is it necessary to have a normed space for the Heine-Borel-Property to hold?

The Heine-Borel-Property says:

A subset M is compact iff it is closed and bounded.

It is well known that the euclidean space $\mathbb{R}^n$ has this property. In a narrower sense I found the following.

A normed space $X$ is finite-dimensional iff it obeys the Heine-Borel-Property.

I was wondering if the property of the space $X$ to have a norm or a metric is necessary. In the proofs I found they are all based on balls, convergence and other notions where norms or metrics are involved. But in the statement of the H-B-Property, all concepts could be defined without a reference to a norm or a metric. For example, let $K$ be a complete, ordered field (like $\mathbb{R}$), then we could define the sets $(a,b) := \{ x : a < x < b \}$ and consider the topology $\tau$ generated by the collection of all such sets. So compactness could be defined as usual by requiring a compact set to have a finite open-cover, boundedness of a set $M$ could be defined as existence of a set $(a,b)$ with $M \subseteq (a,b)$ and closedness as $M$ is closed iff $K \setminus M$ is open, i.e. $K \setminus M \in \tau$. So now could the Heine-Borel-Property be proved for $K$? I don't find any proofs that doesn't use concepts like norm or metric (even the concept of limit could be defined without these concepts, like $x$ is limit point of set $M$ iff for every set $U$ with $x \in U$ if holds that $M \cap U \ne \emptyset$). Do you know any proofs that just use these basic notions?

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You get the same topology if you add sets $(a,+\infty)$ and $(-\infty,b)$ to your basis of $\tau$. This changes the concept of boundedness and hence the validity of Heine-Borel. What makes your chosen basis special? –  Hagen von Eitzen Jan 15 at 18:29
This post may be of interest. –  David Mitra Jan 15 at 18:30
The Schwartz space $\mathcal D(\mathbb R^n)$ of test functions has the Heine-Borel property and is not metrizable.