In topology and mathematics in general, the boundary of a subset of a topological space is the set of points in the closure of not belonging to the interior of . An element of the boundary of is called a boundary point of . The term boundary operation refers to finding or taking the boundary of a set. Notations used for boundary of a set include
\operatorname{bd}(S),\operatorname{fr}(S),
\partialS
Some authors (for example Willard, in General Topology) use the term frontier instead of boundary in an attempt to avoid confusion with a different definition used in algebraic topology and the theory of manifolds. Despite widespread acceptance of the meaning of the terms boundary and frontier, they have sometimes been used to refer to other sets. For example, Metric Spaces by E. T. Copson uses the term boundary to refer to Hausdorff's border, which is defined as the intersection of a set with its boundary.[1] Hausdorff also introduced the term residue, which is defined as the intersection of a set with the closure of the border of its complement.[2]
There are several equivalent definitions for the boundary of a subset
S\subseteqX
X,
\partialXS,
\operatorname{Bd}XS,
\partialS
X
S
S
X
\overline{S}=\operatorname{cl}XS
S
X
\operatorname{int}XS
S
X.
S
p\inX
p
S
S
A boundary point of a set is any element of that set's boundary. The boundary
\partialXS
A connected component of the boundary of is called a boundary component of .
The closure of a set
S
\overline{S}=\operatorname{cl}XS
S
X.
\partialXS~:=~\overline{S}\cap\overline{(X\setminusS)},
\partialXS
X.
("Trichotomy") Given any subset
S\subseteqX,
X
\operatorname{int}XS,\partialXS,
\operatorname{int}X(X\setminusS).
X.
A point
p\inX
p
Conceptual Venn diagram showing the relationships among different points of a subset
S
\Rn.
A
S
B=
S,
S,
S,
S
S
S
A set and its complement have the same boundary:
A set
U
X
\partialXU=X\setminusU.
The interior of the boundary of a closed set is empty.[5] Consequently, the interior of the boundary of the closure of a set is empty. The interior of the boundary of an open set is also empty.[6] Consequently, the interior of the boundary of the interior of a set is empty. In particular, if
S\subseteqX
X
U\subseteq\partialXS
U
X.
A set is the boundary of some open set if and only if it is closed and nowhere dense. The boundary of a set is empty if and only if the set is both closed and open (that is, a clopen set).
Consider the real line
\R
\Q,
\R
\partial(0,5)=\partial[0,5)=\partial(0,5]=\partial[0,5]=\{0,5\}
\partial\varnothing=\varnothing
\partial\Q=\R
\partial(\Q\cap[0,1])=[0,1]
These last two examples illustrate the fact that the boundary of a dense set with empty interior is its closure. They also show that it is possible for the boundary
\partialS
S
X:=\R
\partialS
X
In the space of rational numbers with the usual topology (the subspace topology of
\R
(-infty,a),
a
The boundary of a set is a topological notion and may change if one changes the topology. For example, given the usual topology on
\R2,
\Omega=\left\{(x,y):x2+y2\leq1\right\}
\partial\Omega=\left\{(x,y):x2+y2=1\right\}.
\R3
\Omega=\left\{(x,y,0):x2+y2\leq1\right\},
\partial\Omega=\Omega.
\R2
This example demonstrates that the topological boundary of an open ball of radius
r>0
r
r>0
r
\R2
\R2
X\subseteq\R2
y
Y:=\{0\} x \R
0:=(0,0)\in\R2
X:=Y\cupS1,
\R2
d.
Y,S1,Y\capS1=\{(0,\pm1)\},
\{0\} x [-1,1]
\R2
X.
0=(0,0)
(X,d)
(\R2,d)
Denote the open ball of radius
r>0
(X,d)
Br:=\left\{p\inX:d(p,0)<r\right\}
r=1
y
y=-1
y=1.
(X,d)
r=1
(X,d)
However, the topological boundary
\partialXB1
\operatorname{cl}XB1
X
B1
\partialXB1=\{(0,1),(0,-1)\}
\left\{p\inX:d(p,0)=1\right\}=S1
(X,d).
\operatorname{cl}XB1=B1\cup\{(0,1),(0,-1)\}
\left\{p\inX:d(p,0)\leq1\right\}=S1\cup\left(\{0\} x [-1,1]\right)
(X,d).
(1,0)\inX,
\operatorname{cl}XB1
B1=\{0\} x (-1,1)
X
\{0\} x [-1,1]
\operatorname{cl}XB1.
B1
B1
\partialXB1
\{0\} x [-1,1].
In any metric space
(M,\rho),
M
r>0
c\inM
r
c
Moreover, the unit sphere in
(X,d)
X\setminusY=S1\setminus\{(0,\pm1)\},
X.
\left\{p\inX:d(p,0)=1\right\}
(X,d)
X.
For any set
S,\partialS\supseteq\partial\partialS,
\supseteq
S
S
\partial\partialS=\partial\partial\partialS
S.
In discussing boundaries of manifolds or simplexes and their simplicial complexes, one often meets the assertion that the boundary of the boundary is always empty. Indeed, the construction of the singular homology rests critically on this fact. The explanation for the apparent incongruity is that the topological boundary (the subject of this article) is a slightly different concept from the boundary of a manifold or of a simplicial complex. For example, the boundary of an open disk viewed as a manifold is empty, as is its topological boundary viewed as a subset of itself, while its topological boundary viewed as a subset of the real plane is the circle surrounding the disk. Conversely, the boundary of a closed disk viewed as a manifold is the bounding circle, as is its topological boundary viewed as a subset of the real plane, while its topological boundary viewed as a subset of itself is empty. In particular, the topological boundary depends on the ambient space, while the boundary of a manifold is invariant.
A,
\operatorname{Bdry}(A)
S
X
\overline{S}=S
\partialXS:=\overline{S}\setminus\operatorname{int}XS=S\setminus\operatorname{int}XS.
U
X
U\subseteq\partialXS
U\subseteqS
\partialXS\subseteqS
U\subseteq\operatorname{int}XS
\operatorname{int}XS
X
S
U\subseteq\partialXS=S\setminus\operatorname{int}XS
U\cap\operatorname{int}XS=\varnothing.
U
\operatorname{int}XS
\operatorname{int}XS,
U=\varnothing.
S
X
\partialXS:=\overline{S}\setminus\operatorname{int}XS=\overline{S}\setminusS.
U:=\operatorname{int}X\left(\partialXS\right)
U=\operatorname{int}X\left(\partialXS\right)\subseteq\partialXS=\overline{S}\setminusS,
U\capS=\varnothing.
U ≠ \varnothing
u\inU,
u\inU\subseteq\partialXS\subseteq\overline{S}.
U
u
X
u\in\overline{S},
\overline{S}
U\capS ≠ \varnothing,
\blacksquare
S
X
X\setminusS
X,
\partialXS=\partialX(X\setminusS)
X\setminusS
\operatorname{int}X\partialXS=\operatorname{int}X\partialX(X\setminusS)=\varnothing.
\blacksquare
y
Y=\{0\} x \R
\R2
\R.
\R2\setminusY
\R2.
X
\R2,
X\cap\left(\R2\setminusY\right)=X\setminusY
X.
\blacksquare