Linear continuum

In themathematicalfield oforder theory,acontinuumorlinear continuumis a generalization of thereal line.

Formally, a linear continuum is alinearly ordered setSof more than one element that isdensely ordered,i.e., between any two distinct elements there is another (and hence infinitely many others), andcomplete,i.e., which "lacks gaps" in the sense that everynonempty subsetwith anupper boundhas aleast upper bound.More symbolically:

Shas theleast upper bound property,and
For eachxinSand eachyinSwithx<y,there existszinSsuch thatx<z<y

Asethas the least upper bound property, if every nonempty subset of the set that is bounded above has a least upper bound in the set. Linear continua are particularly important in the field oftopologywhere they can be used to verify whether anordered setgiven theorder topologyisconnectedor not.^[1]

Unlike the standard real line, a linear continuum may be bounded on either side: for example, any (real)closed intervalis a linear continuum.

Examples

The ordered set ofreal numbers,R,with its usualorderis a linear continuum, and is the archetypal example. Property b) is trivial, and property a) is simply a reformulation of thecompleteness axiom.

Examples in addition to the real numbers:

sets which areorder-isomorphicto the set of real numbers, for example a realopen interval,and the same with half-open gaps (note that these are not gaps in the above-mentioned sense)
theaffinely extended real number systemand order-isomorphic sets, for example theunit interval
the set of real numbers with only +∞ or only −∞ added, and order-isomorphic sets, for example ahalf-open interval
thelong line
The setI×I(where × denotes theCartesian productandI= [0, 1]) in thelexicographic orderis a linear continuum. Property b) is trivial. To check property a), we define a map, π₁:I×I→Iby

π₁(x,y) =x

This map is known as theprojection map.The projection map iscontinuous(with respect to theproduct topologyonI×I) and issurjective.LetAbe a nonempty subset ofI×Iwhich is bounded above. Considerπ₁(A). SinceAis bounded above,π₁(A) must also be bounded above. Since,π₁(A) is a subset ofI,it must have a least upper bound (sinceIhas the least upper bound property). Therefore, we may letbbe the least upper bound ofπ₁(A). Ifbbelongs toπ₁(A), thenb×Iwill intersectAat sayb×cfor somec∈I.Notice that sinceb×Ihas the sameorder typeofI,the set (b×I) ∩Awill indeed have a least upper boundb×c',which is the desired least upper bound forA.

Ifbdoes not belong toπ₁(A), thenb× 0 is the least upper bound ofA,for ifd<b,andd×eis an upper bound ofA,thendwould be a smaller upper bound ofπ₁(A) thanb,contradicting the unique property ofb.

Non-examples

The ordered setQofrational numbersis not a linear continuum. Even though property b) is satisfied, property a) is not. Consider the subset

A= {x∈Q|x<√2}

of the set of rational numbers. Even though this set is bounded above by any rational number greater than√2(for instance 3), it has noleast upper boundin the rational numbers.^[2](Specifically, for any rational upper boundr>√2,r/2 + 1/ris a closer rational upper bound; details atMethods of computing square roots § Heron's method.)

The ordered set of non-negativeintegerswith its usual order is not a linear continuum. Property a) is satisfied (letAbe a subset of the set of non-negative integers that is bounded above. ThenAisfiniteso it has a maximum, and this maximum is the desired least upper bound ofA). On the other hand, property b) is not. Indeed, 5 is a non-negative integer and so is 6, but there exists no non-negative integer that lies strictly between them.
The ordered setAof nonzero real numbers

A= (−∞, 0) ∪ (0, +∞)

is not a linear continuum. Property b) is trivially satisfied. However, ifBis the set of negative real numbers:

B= (−∞, 0)

thenBis a subset ofAwhich is bounded above (by any element ofAgreater than 0; for instance 1), but has no least upper bound inB.Notice that 0 is not a bound forBsince 0 is not an element ofA.

LetZ₋denote the set of negative integers and letA= (0, 5) ∪ (5, +∞). Let

S=Z₋∪A.

ThenSsatisfies neither property a) nor property b). The proof is similar to the previous examples.

Topological properties

Even though linear continua are important in the study ofordered sets,they do have applications in the mathematical field oftopology.In fact, we will prove that an ordered set in theorder topologyisconnectedif and only if it is a linear continuum. We will prove one implication, and leave the other one as an exercise. (Munkres explains the second part of the proof in^[3])

Theorem

LetXbe an ordered set in the order topology. IfXis connected, thenXis a linear continuum.

Proof:

Suppose thatxandyare elements ofXwithx<y.If there exists nozinXsuch thatx<z<y,consider the sets:

A= (−∞,y)

B= (x,+∞)

These sets aredisjoint(Ifais inA,a<yso that ifais inB,a>xanda<ywhich is impossible by hypothesis),nonempty(xis inAandyis inB) andopen(in the order topology), and theirunionisX.This contradicts the connectedness ofX.

Now we prove the least upper bound property. IfCis a subset ofXthat is bounded above and has no least upper bound, letDbe the union of allopen raysof the form (b,+∞) where b is an upper bound forC.ThenDis open (since it is the union of open sets), andclosed(ifais not inD,thena<bfor all upper boundsbofCso that we may chooseq>asuch thatqis inC(if no suchqexists,ais the least upper bound ofC), then anopen intervalcontainingamay be chosen that doesn't intersectD). SinceDis nonempty (there is more than one upper bound ofDfor if there was exactly one upper bounds,swould be the least upper bound. Then ifb₁andb₂are two upper bounds ofDwithb₁<b₂,b₂will belong toD),Dand its complement together form aseparationonX.This contradicts the connectedness ofX.

Applications of the theorem

Since the ordered setA= (−∞, 0) U (0,+∞) is not a linear continuum, it is disconnected.
By applying the theorem just proved, the fact thatRis connected follows. In fact anyinterval(or ray) inRis also connected.
The set of integers is not a linear continuum and therefore cannot be connected.
In fact, if an ordered set in the order topology is a linear continuum, it must be connected. Since any interval in this set is also a linear continuum, it follows that this space islocally connectedsince it has abasisconsisting entirely of connected sets.
For an example of atopological spacethat is a linear continuum, seelong line.

References

^Munkres, James (2000).Topology, 2nd ed.Pearson Education.pp. 31, 153.ISBN 0-13-181629-2.
^Hardy, G.H. (1952).A Course of Pure Mathematics, 10th ed.Cambridge University Press.pp. 11–15, 24–31.ISBN 0-521-09227-2.
^Munkres, James (2000).Topology, 2nd ed.Pearson Education. pp. 153–154.ISBN 0-13-181629-2.

[1] Munkres, James (2000).Topology, 2nd ed.Pearson Education.pp. 31, 153.ISBN 0-13-181629-2.

[2] Hardy, G.H. (1952).A Course of Pure Mathematics, 10th ed.Cambridge University Press.pp. 11–15, 24–31.ISBN 0-521-09227-2.

[3] Munkres, James (2000).Topology, 2nd ed.Pearson Education. pp. 153–154.ISBN 0-13-181629-2.

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