{"history":[],"article":{"id":"1ea90573-dd3a-4343-8213-09d978ba0178","title":"Linear continuum","slug":"linear-continuum","content":":::summary\n:::\n\n# \nIn the [mathematical](https://bliptext.com/articles/mathematics) field of [order theory](https://bliptext.com/articles/order-theory), a **continuum** or linear **continuum** is a generalization of the [real line](https://bliptext.com/articles/real-line).\nFormally, a linear continuum is a [linearly ordered set](https://bliptext.com/articles/linearly-ordered-set) S of more than one element that is [densely ordered](https://bliptext.com/articles/dense-order), i.e., between any two distinct elements there is another (and hence infinitely many others), and [complete](https://bliptext.com/articles/completeness-order-theory), i.e., which \"lacks gaps\" in the sense that every [nonempty](https://bliptext.com/articles/empty-set) [subset](https://bliptext.com/articles/subset) with an [upper bound](https://bliptext.com/articles/upper-bound) has a least [upper bound](https://bliptext.com/articles/upper-bound).\nMore symbolically:\n

S has the [least upper bound property](https://bliptext.com/articles/least-upper-bound-property), and

For each x in S and each y in S with x < y, there exists z in S such that x < z < y

\n\nA [set](https://bliptext.com/articles/set-mathematics) has the least upper bound property, if every nonempty sub[set](https://bliptext.com/articles/set-mathematics) of the [set](https://bliptext.com/articles/set-mathematics) that is bounded above has a least upper bound in the [set](https://bliptext.com/articles/set-mathematics).\nLinear continua are particularly important in the field of [topology](https://bliptext.com/articles/topology) where they can be used to verify whether an [ordered set](https://bliptext.com/articles/total-order) given the order [topology](https://bliptext.com/articles/topology) is [connected](https://bliptext.com/articles/connected-space) or not.\nUnlike the standard real line, a linear continuum may be bounded on either side: for example, any (real) [closed interval](https://bliptext.com/articles/closed-interval) is a linear continuum.\n\n# Examples\nExamples in addition to the real numbers:\n\n# Non-examples\n\n# Topological properties\nEven though linear continua are important in the study of [ordered sets](https://bliptext.com/articles/total-order), they do have applications in the mathematical field of [topology](https://bliptext.com/articles/topology).\nIn fact, we will prove that an ordered set in the [order topology](https://bliptext.com/articles/order-topology) is [connected](https://bliptext.com/articles/connected-space) if and only if it is a linear continuum.\nWe will prove one implication, and leave the other one as an exercise.\n(Munkres explains the second part of the proof in )\n**Theorem**\nLet X be an ordered set in the order topology.\nIf X is connected, then X is a linear continuum.\nProof:\nSuppose that x and y are elements of X with x < y.\nIf there exists no z in X such that x < z < y, consider the sets:\nThese sets are [disjoint](https://bliptext.com/articles/disjoint-sets) (If a is in A, a < y so that if a is in B, a > x and a < y which is impossible by hypothesis), [nonempty](https://bliptext.com/articles/nonempty) (x is in A and y is in B) and [open](https://bliptext.com/articles/open-set) (in the order topology), and their [union](https://bliptext.com/articles/union-set-theory) is X.\nThis contradicts the connectedness of X.\nNow we prove the least upper bound property.\nIf C is a subset of X that is bounded above and has no least upper bound, let D be the union of all [open rays](https://bliptext.com/articles/order-topology) of the form (b, +∞) where b is an upper bound for C.\nThen D is open (since it is the union of open sets), and [closed](https://bliptext.com/articles/closed-set) (if a is not in D, then a < b for all upper bounds b of C so that we may choose q > a such that q is in C (if no such q exists, a is the least upper bound of C), then an [open interval](https://bliptext.com/articles/order-topology) containing a may be chosen that doesn't intersect D).\nSince D is nonempty (there is more than one upper bound of D for if there was exactly one upper bound s, s would be the least upper bound. Then if b1 and b2 are two upper bounds of D with b1 < b2, b2 will belong to D), D and its complement together form a [separation](https://bliptext.com/articles/separated-sets) on X.\nThis contradicts the connectedness of X.\n\n## Applications of the theorem","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-26T10:39:33.380Z","updated_at":"2025-01-26T10:39:33.380Z","search_vector":"'continuum':2 'linear':1","search_content":" \n \n\n# \nIn the [mathematical](https://bliptext.com/articles/mathematics) field of [order theory](https://bliptext.com/articles/order-theory), a **continuum** or linear **continuum** is a generalization of the [real line](https://bliptext.com/articles/real-line).\nFormally, a linear continuum is a [linearly ordered set](https://bliptext.com/articles/linearly-ordered-set) S of more than one element that is [densely ordered](https://bliptext.com/articles/dense-order), i.e., between any two distinct elements there is another (and hence infinitely many others), and [complete](https://bliptext.com/articles/completeness-order-theory), i.e., which \"lacks gaps\" in the sense that every [nonempty](https://bliptext.com/articles/empty-set) [subset](https://bliptext.com/articles/subset) with an [upper bound](https://bliptext.com/articles/upper-bound) has a least [upper bound](https://bliptext.com/articles/upper-bound).\nMore symbolically:\n S has the [least upper bound property](https://bliptext.com/articles/least-upper-bound-property), and \n For each x in S and each y in S with x \n \nA [set](https://bliptext.com/articles/set-mathematics) has the least upper bound property, if every nonempty sub[set](https://bliptext.com/articles/set-mathematics) of the [set](https://bliptext.com/articles/set-mathematics) that is bounded above has a least upper bound in the [set](https://bliptext.com/articles/set-mathematics).\nLinear continua are particularly important in the field of [topology](https://bliptext.com/articles/topology) where they can be used to verify whether an [ordered set](https://bliptext.com/articles/total-order) given the order [topology](https://bliptext.com/articles/topology) is [connected](https://bliptext.com/articles/connected-space) or not.\nUnlike the standard real line, a linear continuum may be bounded on either side: for example, any (real) [closed interval](https://bliptext.com/articles/closed-interval) is a linear continuum.\n\n# Examples\nExamples in addition to the real numbers:\n\n# Non-examples\n\n# Topological properties\nEven though linear continua are important in the study of [ordered sets](https://bliptext.com/articles/total-order), they do have applications in the mathematical field of [topology](https://bliptext.com/articles/topology).\nIn fact, we will prove that an ordered set in the [order topology](https://bliptext.com/articles/order-topology) is [connected](https://bliptext.com/articles/connected-space) if and only if it is a linear continuum.\nWe will prove one implication, and leave the other one as an exercise.\n(Munkres explains the second part of the proof in )\n**Theorem**\nLet X be an ordered set in the order topology.\nIf X is connected, then X is a linear continuum.\nProof:\nSuppose that x and y are elements of X with x x and a < y which is impossible by hypothesis), [nonempty](https://bliptext.com/articles/nonempty) (x is in A and y is in B) and [open](https://bliptext.com/articles/open-set) (in the order topology), and their [union](https://bliptext.com/articles/union-set-theory) is X.\nThis contradicts the connectedness of X.\nNow we prove the least upper bound property.\nIf C is a subset of X that is bounded above and has no least upper bound, let D be the union of all [open rays](https://bliptext.com/articles/order-topology) of the form (b, +∞) where b is an upper bound for C.\nThen D is open (since it is the union of open sets), and [closed](https://bliptext.com/articles/closed-set) (if a is not in D, then a < b for all upper bounds b of C so that we may choose q > a such that q is in C (if no such q exists, a is the least upper bound of C), then an [open interval](https://bliptext.com/articles/order-topology) containing a may be chosen that doesn't intersect D).\nSince D is nonempty (there is more than one upper bound of D for if there was exactly one upper bound s, s would be the least upper bound. Then if b1 and b2 are two upper bounds of D with b1 < b2, b2 will belong to D), D and its complement together form a [separation](https://bliptext.com/articles/separated-sets) on X.\nThis contradicts the connectedness of X.\n\n## Applications of the theorem","revisionCount":0},"hasMore":false,"nextCursor":null}

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