The already given answer and comments have essentially solved the problem; however I

will show the way the question can be studied from the point of view of

"classical" lattice theory (say, the times of von Neumann and

Birkhoff, then Halperin, Kaplansky, F. and M. Maeda)

For complete lattices, it is well known that algebraic implies

Scott-continuous implies meet-continuous.

{Remember the

definitions: a complete lattice is a partial order where every subset

has a supremum (equivalently, every subset has a infimum). It is

meet-continuous when *increasing* joins distribute over meet

(dually for join continuity); it is Scott continuous when every element

$a$ is the join of elements $b$ "way below" $a$ (meaning: every

increasing join that is at least $a$ has a member that is at least $b$);

it is algebraic when meet-continuous and each $a$ is the join of

join-inaccessible elements $b$ (elements that are "way below"

themselves).}

The term "continuous" tout-court might seem suitable for

Scott-continuity, since the latter coincides with embeddability (with

preservation of all meets and increasing sups) in a power of the real

interval $[0,1]$ (the "archetipe" of the continuum), in the same ways

as algebraicity uses the two-element chain $\{0,1\}$ (the "archetipe"

of discreteness) instead of $[0,1]$. However, this also means that such

a concept of continuity is a generalization of (and not a opposition to)

discreteness.

For classical objects of lattice theory (like complemented modular

lattices or orthomodular lattices) which are related to (discretely or

continuously valued) dimension functions, the term "continuity" is

traditionally associated with meet-continuity and its dual (von

Neumann's [meet and/or join] "continuous geometries").

{For the concept of semiorthogonality (used below) and related

ones, see S. Maeda papers, especially the last paper (1961, freely

available inside projecteuclid) about dimension lattices.}

The point (to be proved below): for such classical (complete, relatively

semi-orthocomplemented) lattices, Scott-continuity coincides with

algebraicity and even "compactly atomistic" (hence usually even more:

direct product of "finite discrete factors": the discrete subcase of

continuous geometries).

The lemma: in a complete (relatively) semiorthocomplemented lattice,

"$b$ way below $a$" implies "$b$ join inaccessible" (hence compact

in the meet-continuous case).

Proof: suppose $b$ inceasing join of the $b_i$. Fix a

semi-orthocomplement $c$ of $b$ in $a$; then $b_i\oplus c$ is a

increasing family, with join $a$ (it contains $c$ and the join $b$ of

the $b_i$; conversely, each $b_i\oplus c$ is contained in $b\oplus
c=a$). By definition of "$b$ way below $a$", $\exists i:b_i\oplus
c\geq b$. Adding $c$, $b_i\oplus c\geq b\oplus c$. Now, write

$b=b_i\oplus c_i$; so: $b_i\oplus c\geq b_i\oplus c_i\oplus c$; hence:

$c_i=0$, $b_i=b$.

From the lemma it follows that a Scott-continuous relatively

semiorthocomplemented lattice is algebraic (it is meet-continuous, hence

"$b$ way below $a$" implies "$b$ compact" and so each element is

join of compact elements); then it is also atomistic (i.e. sectionally

semicomplemented and atomic: use relatively complemented [whci follows

from semi-ortho-complemented] and weakly atomic [which follows from

algebraic]: every interval $[a,b]$ contains a covering $\{a',b'\}$ and

the realtive orthocompement $c'$ of $a'$ in $b'$ is an atom in $[a.b]$:

if not, $c'$ properly splits in $x\oplus y$ and so between $a'$ and $b'$

one has $a'\oplus x$).

Once one has a compactly atomistic lattice, a weak (semimodularity)

condition (like the covering property, which is satisfied by

meet-continuous geometries and projection ortholattices of

AW$^*$-algebras and their Jordan analogues) is well known to imply a

decomposition into a direct product of subdirectly irriducible factors

of the same kind (see "matroid lattices" in F. Maeda - S. Maeda book,

theory of symmetric lattices).

In the semimodular and orthomodular case (i.e. dimension ortholattices),

the irreducible factors are exactly the finite-dimensional

orthocomplemented projective geometries; conclusion: Scott-continuous

semimodular otholattices are exactly the direct products of

(irreducible) orthocomplemented finite-dimensional projective

geometries. [The W$^*$ or AW$^*$ algebras that give such ortholattices

are exactly the direct products, in the category of such algebras, of

matrix $*$-algebras over the real, complex or quaternional $*$-field.]

{The direct product as (algebras or) rings (with involution) of

matrix $*$-algebras with scalar entries is a $*$-regular ring hence it

is not even a C$^*$-algebra (unless there are only finitely many

nontrivial factors); the direct product in the suitable category is

smaller (one takes only the bounded elements of the $*$-regular ring,

see Berberian's book "Baer$^*$-rings").}

In the complemented modular case (i.e. meet-continuous geometries), the

irreducible factors are exactly the (possibly infinite dimensional)

projective geometries (associated to vector spaces over a sfield, except

possibly nonarguesian planes, and lines and points). Conclusion: the

Scott-continuous complemented modular lattices are exactly the (possibly

reducible) projective geometries; subcase: the Scott-continuous

continuous geometries are exactly the direct products of

finite-dimensional projective geometries.

Note: this generalizes the characteriazion of Scott-continuous boolean

algebras in the "compendium of continuous lattices".

Compendium on continuous lattices, continuous lattices form an overlapping subclass of continuous geometry. – Rennela Jun 10 '13 at 9:28