# Is a closed connected semilattice of $C(I)$ path-connected?

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 $$0\le i and since for $$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 $$s in $$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.

• Could you please elaborate on why connectedness of a closed sublattice imply path-connectedness?
– erz
Jun 11 '20 at 3:25
• Sure; I've edited and added a proof Jun 11 '20 at 8:15
• 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)
– erz
Jun 11 '20 at 23:12
• 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. Jun 12 '20 at 5:32