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In general, a topology is not determined by its compact sets, meaning there are topological spaces $(X, \tau_1)$, $(X, \tau_2)$ such that $\tau_1\ne \tau_2$ and still $K\subseteq X$ is compact with respect to $\tau_1$ if, and only if it is compact with respect to $\tau_2$. I wonder if there is such a compact-twin topology for the real numbers and its powers.

In other words, let $\tau$ be a topology over $\mathbb R^n$ so that $K\subseteq \mathbb R^n$ is compact if, and only if, $K$ is closed and bounded (with respect to the usual metric). Must $\tau$ be the usual topology?

If we assume compact sets are closed in $\tau$ (this is true, for instance, if $\tau$ is Hausdorff) then the usual open neighborhoods of $\infty$ belong to $\tau$. However, those cannot be the only open sets in $\tau$, since otherwise we'd have unbounded compact sets.

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    $\begingroup$ Some analyst is going to march in here and tell us that the answer depends on whether or not you accept AoC. $\endgroup$ Commented yesterday
  • $\begingroup$ Indeed, any closed set in the usual topology is compact in the topology of neighborhoods of $\infty.$ $\endgroup$ Commented yesterday
  • $\begingroup$ There is certainly no strictly finer topology on $\mathbb R^n$ with the same compact sets. $\endgroup$ Commented yesterday

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Define $\tau'$ as the set of complements of the union of a compact subset and a closed discrete subset of $\mathbb R$. This is a coarser topology than the usual (noted $\tau$).

So every $\tau$-compact subset is necessarily $\tau'$-compact.

EDIT: Thanks to Sassatelli Giulio's comment, the converse is now complete.

Let $K$ be a $\tau'$-compact set and assume by contradiction $K$ is unbounded. Then there is a sequence $(x_n)$ of points in $K$ such that for all $n$, $|x_n|>n$, and $\bigcup_n(\mathbb R\setminus\{x_n,x_{n+1},\dots\})$ is an open cover of $K$ without finite subcover, which is absurd.

Therefore every $\tau'$-compact set is bounded, and the topologies $\tau$ and $\tau'$ agree over an open ball, so every $\tau'$-compact is $\tau$-compact.

Finally, compact subsets are the same under both topologies.

Note that the co-compact topology cannot be sufficient to the OP, because the whole space would be compact.

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  • $\begingroup$ @Ulli: If all unions of a compact and finite set are closed, then some sets that are not of the form "union of a compact and a closed discrete set" are also closed: For $r>0$, consider $A_r=[-r,r]\cup\{\,1/n:n\in\mathbb N, n<1/r\,\}$. Their intersection is $\{0\}\cup\{\,1/n:n\in\mathbb N\,\}$. -- Can the idea be saved by something like closed = "union of a compact set $C$ and a discrete set $D$ such that $\overline D\setminus D\subset C$"? Things are beginning to get complicated ... $\endgroup$ Commented yesterday
  • $\begingroup$ Come to think of it, why not force $\tau'$ to be coarser than $\tau$ by letting $\tau'$ be the set of all $\tau$-open sets whose complement is the union of a $\tau$-compact and a $\tau$-discrete set? $\endgroup$ Commented yesterday
  • $\begingroup$ @HagenvonEitzen $\bigcap A_r$ is $\tau$-compact. And your last proposal is identical to mine. $\endgroup$ Commented yesterday
  • $\begingroup$ Why must $\tau'$-compact subsets be $\tau'$-closed? $\tau'$ is not T2 (in point of fact, I'm quite confident that it's an irreducible topology). $\endgroup$ Commented yesterday
  • $\begingroup$ I was so confident I had written "closed" before "discrete" that I didn't saw it still missed. $\endgroup$ Commented yesterday
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Here is a $T_2$-example:

$\tau$ denotes the usual topology on $\mathbb R$. Let $\mathcal U$ be a free ultrafilter on $\mathbb N$.
We define a topology $\sigma$ on $\mathbb N_0$: Each $n \in \mathbb N$ is isolated, and neighborhoods of $0$ are of the form $\{0\} \cup U, \space U \in \mathcal U$, i.e., $(\mathbb N_0, \sigma)$ is a subspace of $\beta \mathbb N$ with $0 \in \beta \mathbb N \setminus \mathbb N$.

Now define $\rho = \{U: U \in \tau, U \cap \mathbb N_0 \in \sigma\}$. It is easy to see that $\rho$ is a topology on $\mathbb R$. @Jakobian defined this kind of topology in their answer here, and proved that it is $T_2$ (non-trivial!).

Of course, $\rho$ is strictly coarser than $\tau$ [$(-1,1)$ is $\tau$-open, but not $\rho$-open]. Hence, $\tau$-compact subsets of $\mathbb R$ are $\rho$-compact. But also vice versa:

Let $A$ be a $\rho$-compact subset of $\mathbb R$. Since $\rho$ is $T_2$, $A$ is closed w.r.t. to $\rho$ and $\tau$. Assume $A$ is unbounded w.r.t. the usual metric on $\mathbb R$.

Case 1: $A \setminus \mathbb N_0$ is unbounded above. By induction, choose $u_n \in \mathbb N, x_n \in A$ such that $n < u_n < x_n < u_n + 1 < u_{n+1}$ for each $n \in \mathbb N$.
Define $B = \{x_n: n \in \mathbb N\} \subseteq A$. We have $(u_n, u_n+1) \in \rho$ and $B \cap (u_n, u_n+1) = \{x_n\}$, hence $B$ is discrete w.r.t. $\rho$. Moreover, $B$ is closed w.r.t. $\tau$, $B \cap \mathbb N_0 = \emptyset$, hence $B$ is closed w.r.t. $\rho$, contradicting compactness of $A$.

Case 2: $A \setminus \mathbb N_0$ is unbounded below. Then define a strictly decreasing sequence $(x_n)_{n \in \mathbb N} \subseteq A \cap \mathbb R^{< 0}$, which is unbounded below, and argue similar to case 1.

Case 3: $A \setminus \mathbb N_0$ is bounded. Since $\mathbb N_0$ is $\rho$-closed, $B = A \cap \mathbb N_0$ is infinite and compact w.r.t. $\rho$. Since $\mathbb N_0$ is $\tau$-discrete, $\rho|\mathbb N_0 = \sigma$, hence $B$ is compact w.r.t. $\sigma$, contradicting that each infinite, compact subspace of $\beta \mathbb N$ is uncountable.

Remark
Using normality of $\tau$ and of $\sigma$, it can also be shown that $\rho$ is normal. This provides an alternative proof of $\rho$ being $T_2$, since it is rather easy to see that $\rho$ is $T_1$. (See the remark in the above referenced answer by Jakobian.)

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