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The quotient map is open, and every interval shorter than one embeds in R/Z

Statement

Let p:RR/Z be the quotient map of The circle as S1=R/Z with basepoint [0]. The quotient map is open, and every interval shorter than one embeds in R/Z.

More precisely, for every open UR,

p1(p[U])=nZ(U+n),

and p[U] is open. If ab, ba<1, and J is any of (a,b), [a,b], [a,b), or (a,b], then pJ:Jp[J] is a homeomorphism, with both sides carrying their subspace topologies.

Facts & Assumptions

Given: The quotient projection p, an open set UR, and an interval J of one of the displayed four forms with length =ba<1.

[L1]

Let p:RR/Z be the quotient projection inducing the quotient topology, with p(x)=[x] and p(x)=p(y) exactly when xyZ (The circle as S1=R/Z with basepoint [0]).

[L2]

Identify Z with its canonical copy inside R. Then for every real x there is exactly one integer m with mx<m+1 (Integer part: for every real x there is exactly one integer m with mx<m+1).

[L3]

A function f:XY is an open map if f[V] is open in Y for every open VX; an embedding is a homeomorphism onto its image with the subspace topology (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L4]

Every constant real-valued function and the identity are continuous, and finite sums and scalar multiples of continuous real-valued functions are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

Proof

technique · direct
1.1

A real x lies in p1(p[U]) exactly when p(x)=p(u) for some uU, which by [L1] is equivalent to xu=n for some nZ; hence p1(p[U])=nZ(U+n). Each translate U+n is open: translating by n and by n gives mutually inverse continuous maps by [L4]. The union is open, so the quotient-topology criterion in [L1] makes p[U] open. Thus p is open in the sense of [L3], including when U=.

L1L3L4
1.2

Suppose x,yJ and p(x)=p(y). Then k:=xyZ by [L1], while k=xy<1. If 0<k<1, both 0 and k satisfy the integer-part inequalities for the real k, contrary to uniqueness in [L2]; applying the same argument to k excludes 1<k<0. Hence k=0 and x=y, so pJ is injective. This also covers a singleton interval; for an empty interval injectivity is vacuous.

L1L2algebra
2.1

The restriction pJ is continuous and is a bijection onto p[J] by step 1.2. To prove its inverse continuous, let O be relatively open in J and xO. Choose δ>0 with J(xδ,x+δ)O, and put r=12min{δ,1}>0. Step 1.1 makes W:=p[(xr,x+r)] open. If zJ and p(z)W, choose y(xr,x+r) with p(z)=p(y); then zyZ by [L1] and zyzx+xy<+r<1, so [L2] gives z=y. Thus zJ(xδ,x+δ)O, and Wp[J]p[O]. Every point of p[O] therefore has a relative open neighbourhood contained in p[O], so p[O] is open in p[J]. The empty case has the unique empty inverse. Hence pJ is a homeomorphism onto its image, and therefore an embedding by [L3].

step 1.1step 1.2L1L2L3L4algebra

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