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.

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  • $\begingroup$ 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? $\endgroup$ – მამუკა ჯიბლაძე May 5 at 4:50
  • $\begingroup$ @მამუკაჯიბლაძე X is assumed to be locally path connected (perhaps), likely one may do the same with a weaker assumption. $\endgroup$ – Anton Petrunin May 5 at 5:01
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    $\begingroup$ @მამუკაჯიბლაძე local path connected + connected implies path connected: the path components are open and disjoint, and therefore there is just one such path component $\endgroup$ – erz May 5 at 6:37
  • $\begingroup$ 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? $\endgroup$ – erz May 5 at 6:55
  • $\begingroup$ @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. $\endgroup$ – 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. $\rho(x,y)=r<R<1$. 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 $\rho(x,z)=\frac{r+R}{2}<R$. 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 $X=\{(t,\sin \frac{1}{t}), 0<t\le 1\}\cup\{(0,0\}$, 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.

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