# Is it possible to connect every compact set?

Let $$X$$ be a "nice" space: metrizable, connected, locally path connected perhaps. Let $$K\subset X$$ be a compact set.

Is there a always a compact connected $$L\subset X$$ such that $$K\subset L$$?

This is true if we assume local compactness: cover $$K$$ with a finite number of connected relatively compact open sets, take their closure, and then join with arcs. However, without local compactness I don't know what to do.

Choose a sequence $$\varepsilon_n\to 0$$ and a $$\varepsilon_n$$-net $$N_n$$ for each $$n$$. Assume $$N_0$$ is a one-point set. For each point in $$x\in N_k$$ choose a closest point in $$y\in N_{k-1}$$ and connect $$x$$ to $$y$$ by a curve. Note that we can assume that diameter of the curve is at most $$\delta_k$$ for a fixed sequence $$\delta_k\to 0$$.

Consider the union $$K'$$ of all these curves with $$K$$; observe that $$K'$$ is compact and path connected.

• Why is it possible to connect $x$ to $y$? Are you assuming path connectedness of $X$? I think the assumption is connected but only locally path connected? – მამუკა ჯიბლაძე May 5 at 4:50
• @მამუკაჯიბლაძე X is assumed to be locally path connected (perhaps), likely one may do the same with a weaker assumption. – Anton Petrunin May 5 at 5:01
• @მამუკაჯიბლაძე local path connected + connected implies path connected: the path components are open and disjoint, and therefore there is just one such path component – erz May 5 at 6:37
• Do I understand correctly that the control of the diameters of the curves comes from the fact that any locally path connected metrizable space admits a metric such that every ball of diameter less than some fixed number is path connected? Or is there a simpler way? – erz May 5 at 6:55
• @erz, yes, it can be done this way, but this theorem is hard (and I do not know its proof). Instead one may directly apply existence of arbitrary small path connected neighborhood. – Anton Petrunin May 5 at 19:23

This is meant to fill in some of the details outlined by Anton Petrunin's answer, and also to refine the statement slightly. Recall that a compact connected Hausdorff space is called a continuum.

We will call a topological space $$X$$ continuum-connected if every $$x,y\in X$$ can be joined by a continuum, i.e. there is a continuum $$K\subset X$$ that contains both $$x$$ and $$y$$. We will call $$X$$ locally continuum-connected if for every $$x\in X$$ and open neighborhood $$U$$ of $$x$$ there is an open neighborhood $$V$$ of $$x$$ such that every $$y\in V$$ can be joined by a continuum within $$U$$. It is easy to see that continuum-components of locally continuum-connected space are open and disjoint, and so a connected locally continuum-connected is continuum-connected.

Proposition. A metrizable space $$X$$ is locally continuum-connected if and only if there is a metric $$\rho$$ on $$X$$ compatible with the topology and such that every open ball of radius less than $$1$$ is continuum-connected.

This is analogous to Theorem IV.7.1 in Newman - Elements of the topology of plane sets of points. There it is stated for (locally) connected metrizable spaces, but works also for any (locally) set-connected metrizable spaces, for an appropriate collection of connected sets (e.g. separable, bounded, arcs).

Proof. Sufficiency is clear. Let us prove necessity. Choose an arbitrary metric $$d$$ on $$X$$ bounded by $$1$$. For $$x,y\in X$$ declare $$\rho(x,y)$$ to be the infimum of diameters of the continuums that join $$x$$ and $$y$$ (if $$x$$ and $$y$$ are not joined by any continuum put $$\rho(x,y)=1$$). It is easy to see that $$\rho$$ is a metric, and moreover $$d\le\rho$$. Furthermore, if $$x_n\to x$$, since $$X$$ is locally continuum-connected, $$x_n$$ and $$x$$ can be joined by arbitrarily small continuums, and so $$\rho(x_n,x)\to x$$. Thus, $$\rho$$ is equivalent to $$d$$, and so is compatible with the topology of $$X$$.

It is left to show that every ball of radius less than $$1$$ is continuum-connected. Let $$x\in X$$ and let $$R<1$$. Assume that $$y\in B_{\rho}(x,R)$$, i.e. $$\rho(x,y)=r. By definition of $$\rho$$, there is a continuum $$K$$ with $$d$$-diameter at most $$\frac{r+R}{2}$$ that joins $$x$$ and $$y$$. Every point $$z\in K$$ is joined with $$x$$ by $$K$$, and so $$\rho(x,z)=\frac{r+R}{2}. Hence, $$K\subset B_{\rho}(x,R)$$, and so $$y$$ is joined by $$x$$ by a continuum in $$B_{\rho}(x,R)$$. $$\square$$

Corollary. A metrizable space $$X$$ is locally continuum-connected if and only if every point has a base of open continuum-connected neighborhoods.

Now, having these characterizations we can answer the original question.

Theorem. Let $$X$$ be a connected and locally continuum-connected metrizable space. Then for every compact $$K\subset X$$ there is a continuum $$L\subset X$$ that contains $$K$$.

Before proving the theorem, let us prove the following characterization of compactness.

Lemma Let $$Y$$ be a metric space for which there is a compact $$K\subset Y$$ such that for every $$\varepsilon>0$$ there is a compact $$N$$ such that $$K$$ is an $$\varepsilon$$-net of $$Y\backslash N$$. Then $$Y$$ is compact.

Proof. It is clear that $$Y$$ is completely bounded. We only need to prove completeness. Let $$\{y_m\}\subset Y$$ be a Cauchy sequence. It is enough to find a convergence subsequence. For every $$k$$ let $$N_k$$ be compact and such that $$K$$ is $$\frac{1}{k}$$-net for $$Y\backslash N_k$$. We may assume that $$N_k\subset N_{k+1}$$.

If an infinite subsequence of $$\{y_m\}$$ was contained in $$N_k$$, for some $$k$$, then there would be a convergent subsequence due to compactness of $$N_k$$. Hence, we can choose a subsequence $$\{z_m\}$$ such that $$z_m\not\in N_m$$. Since $$K$$ is an $$\frac{1}{m}$$-net for $$Y\backslash N_m$$, there is $$x_m\in K$$ with $$\rho(x_m,z_m)<\frac{1}{m}$$. Since there is a subsequence of $$\{x_{m_k}\}$$ that converge to $$x\in K$$, so does $$\{z_{m_k}\}$$. $$\square$$

Proof of the theorem. Using the proposition, we can metrize $$X$$ with a metric such that open balls of radius less than $$1$$ are continuum-connected.

For natural $$n$$, let $$K_n\subset K$$ be a finite $$\frac{1}{2^n}$$-net of $$K$$. For every $$x\in K_{n+1}$$ there is $$y\in K_{n}$$ such that $$\rho(x,y)<\frac{1}{2^n}$$. Since $$B(y, \frac{1}{2^n})$$ is continuum-connected, there is a continuum $$L^n_{x}\subset B(y, \frac{1}{2^n})$$. Then for any $$m>n$$ and $$x\in K_m$$ and $$z\in L_x$$ there $$y\in K_{n}$$ such that $$\rho(z,y)<\frac{1}{2^{n-1}}$$.

Let $$z\in K$$ and for $$x\in K_1$$ let $$L^1_x$$ be a continuum that joins $$x$$ with $$z$$. Observe by induction that $$M_n=\bigcup_{i\le n, x\in K_n} L_{x}^i$$ is a continuum, and so $$M= \bigcup M_k$$ is connected. Since $$M$$ contains an $$\frac{1}{2^n}$$-net of $$K$$, for every $$n$$, it follows that $$K\subset \overline{M}$$. Hence, $$M\subset M\cup K\subset \overline{M}$$ from where $$Y=M\cup K$$ is connected.

Finally, since $$K_n\subset K$$ is a $$\frac{1}{2^{n-1}}$$-net for $$K\cup \bigcup_{k>n} M_k\supset Y\backslash M_n$$, for every $$n$$, $$Y$$ is compact due to the Lemma.$$\square$$

Remark. I also would like to present a nice example that bof gave in the comments (now deleted), that at least local connectedness is required: Consider the following modification of the topologist's sine curve $$X=\{(t,\sin \frac{1}{t}), 0, which is connected and moreover is a polish space. However the compact set $$\{(x,y)\in X, y=0\}$$ cannot be connected by a continuum.