4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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-wise 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-wise 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-wise 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-wise 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-wise 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-wise 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-wise connected if and only if every point has a base of open continuum-wise connected neighborhoods.

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

Theorem. Let $X$ be a connected and locally continuum-wise 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-wise 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-wise 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. Note that for a completely metrizable space local connectedness is equivalent to local path-connectedness.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The usual term you will find in the literature is "continuum-wise connected". I have also seen "semi-continuum" or "semicontinuum" to refer to such a space. $\endgroup$ – D.S. Lipham Jun 8 at 21:09
  • $\begingroup$ @D.S.Lipham thank you! now fixed $\endgroup$ – erz Jun 9 at 4:37
0
$\begingroup$

Here is another answer, based again on Anton Petrunin's idea, but obtaining a slightly different result.

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

In order to prove this result we need a version of the proposition from my previous answer.

Proposition. A locally path-connected completely metrizable space $X$ supports a complete metric $\rho$ on $X$ compatible with the topology and such that every open ball of radius less than $1$ is path-connected.

Proof. Let $d$ be a complete metric on $X$ bounded by $1$. Apply the same construction as in my previous answer (but with paths instead of continuums) and obtain $\rho$. Since $\rho\ge d$ are equivalent, and the latter is complete it is easy to see that the former is also complete (a $\rho$-Cauchy sequence is a $d$-Cauchy sequence, hence it is $d$-convergence, and so $\rho$-convergent).$\square$

Proof of the Theorem. We will construct a convergent sequence of paths $\varphi_n:[0,1]\to X$ such that the image of $\gamma_n$ contains a $\frac{1}{2^n}$-net of $K$.

For natural $n$, let $K_n\subset X$ be a finite $\frac{1}{2^n}$-net of $K$. Using connectedness one can choose them so that $K_n\cap K_m=\varnothing$. Moreover, let $K_1=\{x_0,...,x_n\}$.

Let $\gamma_1:[0,1]\to X$ be a continuous path such that $\gamma_1|_{[\frac{2i}{2n+1},\frac{2i+1}{2n+1}]}\equiv x_i$, $i=0,...,n$ (on the intermediate segments $\gamma_1$ joins $x_i$ with $x_{i+1}$, which is possible since $X$ is path connected).

For $0\le a<b\le 1$ and $x,y\in X$ with $\rho(x,y)<r<1$ let $\gamma:[a,b]\to X$ be a a continuous loop such that $\gamma(a)=\gamma(b)=x$, $\gamma|_{[\frac{2a+b}{3},\frac{a+2b}{3}]}\equiv y$, and the image of $\gamma$ is contained in $B(x,r)$ (which is possible since open balls of radius less than $1$ are path connected).

Now assume that $\gamma_n$ is constructed so that its image contains $K_n$ and for every $x\in K_n$ there are $a<b$ such that $[c,d]\subset \gamma^{-1}_n(x)$. Let $x_1,...,x_m\in K_{n+1}$ be such that $\rho(x_i,x)<\frac{1}{2^n}$. Re-define $\gamma_n$ on $[c,d]$ to be a series of loops defined above from $x$ to $x_1$ and back, then from $x$ to $x_2$ and back, and so on.

Applying the same construction to all elements of $K_n$ (simultaneously) we get $\gamma_{n+1}$ such that for every $y\in K_{n+1}$ there are $a<b$ such that $[a,b]\subset \gamma^{-1}_{n+1}(y)$. Moreover, if $\gamma_{n+1}(t)\ne \gamma_n(t)$, it follows that $\gamma_{n+1}(t)\in B(\gamma_n(t),\frac{1}{2^n})$, from where $\rho(\gamma_{n+1},\gamma_n)\le \frac{1}{2^n}$.

Note that the image of $\gamma_{n+1}$ contains the image of $\gamma_n$. Moreover, from construction and the fact that $K_m$'s are disjoint that if $x\in K_n$, then $\gamma_{n+1}^{-1}(x)\ne \varnothing$ and if $t\in \gamma_{n+1}^{-1}(x)$, then $\gamma_m(t)=x$, for all $m>n$.

It follows that $\{\gamma_n\}$ is a Cauchy sequence of maps from $[0,1]$ into a complete space $X$. Consequently, it uniformly converges to a map $\gamma:[0,1]\to X$. From the previous paragraph it follows that the image of $\gamma$ contains every $K_n$, and since it is compact, it contains $\overline{\bigcup K_n}\supset K$. $\square$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.