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Let $\Gamma $ be a sub-lattice of the Banach space $\big( B(S),\|\cdot\|_\infty\big)$ of all bounded real valued functions on the set $S$ (meaning that for any $f,g\in\Gamma $ both functions $f\wedge g: x\mapsto \min(f(x),g(x))$ and $f\vee g: x\mapsto\max(f,g)$ belong to $\Gamma$).

Then, it is not hard to see that if $\Gamma$ is closed and connected, it is also path-connected (details below).

But what if $\Gamma$ is only assumed to be a closed $\wedge$-semilattice, that is, $f\wedge g\in\Gamma$ for any $f,g\in\Gamma$. Is it still true that $\Gamma$ is path-connected? Additional hypotheses may be assumed, e.g. $S$ a topological space and $\Gamma\subset C_b(S)$.

Question: Is a closed connected $\wedge$-semilattice $\Gamma$ of $C([0,1])$ path-connected? And if $\Gamma$ is also compact?

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Details. Proof that a closed connected lattice $\Gamma\subset B(S)$ is path-connected (in fact, even locally path connected).

Recall that, as a general fact, in a metric space $(\Gamma,d)$, any pair of elements $f$ and $g$ of $\Gamma$, for any $\epsilon>0$, are connected by $\epsilon$-chains. That means: for any $\epsilon>0$ there is $m\in\mathbb{N}$ and a finite sequence $(g_i)_{0\le i \le m}\subset \Gamma$ such that $g_0=g$, $g_m=f$ and $d(g_{i+1},g_i)<\epsilon$ for $i=0,\dots,m-1$. (This because the set of $f$ that can be connected to $g$ by an $\epsilon$-chain is a non-empty closed and open set, thus the whole space $\Gamma$).

In the case of a lattice $\Gamma\subset B(S)$, assuming $g\le f$, the $\epsilon$-chain can be taken increasing w.r.to the point-wise order: just replace $g_i$ with $g^*_i:=(g_0\vee\dots\vee g_i)\wedge f\in\Gamma$: then clearly $g^*_0=g$, $g^*_m=f$, $g^*_i\le g^*_{i+1}$ for $0\le i<m$ and since for $h\in\Gamma$ the maps $\Gamma\ni u\mapsto u\vee h\in \Gamma$ and $\Gamma\ni u\mapsto u\wedge h\in \Gamma$ are $1$-Lip, it is also true $$\|g^*_{i+1}-g^*_i\|_\infty=\big\|\big((g_0\vee\dots\vee g_i)\vee g_{i+1})\wedge f-\big((g_0\vee\dots\vee g_i)\vee g_i\big)\wedge f\big\|_\infty $$ $$\le\|g_{i+1}-g_i\|_\infty<\epsilon.$$ Moreover, we may re-indicize monotonically $\epsilon$-chains on finite subsets of $[0,1]$. If we iterate this construction, between consecutive elements of an $\epsilon$-chain, with $\epsilon=1/k$ for $k=0,1,2,\dots$, we precisely get by induction:

A nested sequence of finite sets $J_0:=\{0,1\}\subset J_1 \subset\dots\subset J_k\subset J_{k+1}\subset\dots\subset [0,1]$, with $J:=\cup_{k\ge0}J_k$ dense in $[0,1]$, and a sequence of increasing maps $\alpha_k:J_k\to\Gamma$ such that $\alpha_0(0)=g$, $\alpha_o(1)=f$, ${\alpha_{k+1}}_{|J_k}=\alpha_k$, $\|\alpha(s')-\alpha(s)\|_\infty<1/k$ for any pair of consecutive elements $s,s'$ of $J_k$.

Therefore the maps $\alpha_k$ glue to an increasing map $\alpha:J\to \Gamma$, and I claim this map is uniformly continuous. Indeed for any $k\ge1$ and $t,t'\in J$ with $|t-t'|\le\delta_k$, where $$ \delta_k:=\min\{|s-s'|: s,s'\in J_k, s\ne s'\}>0$$ there are consecutive points $s<s'<s''$ in $J_k$ such that $s'\le t,t'\le s''$, so that by monotonicity of $\alpha$ $$\|\alpha(t')-\alpha(t)\|_\infty\le\|\alpha(s'')-\alpha(s)\|_\infty\le2/k.$$ So $\alpha$ is uniformly continuous and extends by density to a continuous path in $\Gamma$ between $g$ and $f\ge g$. For the general case, we may juxtapose a path from $g$ to $g\vee f$ and one from $g\vee f$ to $f$, showing that $\Gamma$ is path connected.

Notice that if $f$ is $\epsilon$-close to $g$, so is the whole path, so $\Gamma$ is indeed also locally path connected.

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    $\begingroup$ Could you please elaborate on why connectedness of a closed sublattice imply path-connectedness? $\endgroup$
    – erz
    Jun 11 '20 at 3:25
  • $\begingroup$ Sure; I've edited and added a proof $\endgroup$ Jun 11 '20 at 8:15
  • $\begingroup$ so you proved the following: if $\Gamma$ is a lattice, endowed with a complete metric that makes $f\to f\wedge g$ and $f\to f\vee g$ $1$-Lipschitz, then any $f\ge g$ can be joined by a monotone continuous path within $\overline{B}(f, d(f,g))$. And in the process you also showed that a monotone map into such lattice from a dense subset of $[0,1]$ is uniformly continuous as long as there are $\varepsilon$ stairs for every $\varepsilon$. Interesting! (sorry for an unhelpful comment) $\endgroup$
    – erz
    Jun 11 '20 at 23:12
  • $\begingroup$ It seem that the map $C[0,1]\to C[0,1]$, $f\mapsto- f$, is an isomorphism of $C[0,1]$ endowed with the $\vee$-semilattice operation to $C[0,1]$ endowed with the $\wedge$-semilattice operation. So, these topological semilattices are isomorphic and should have the same properties. $\endgroup$ Jun 12 '20 at 5:32

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