Boundary (topology)

A set and its complement have the same boundary:

$${\displaystyle \partial _{X}S=\partial _{X}(X\setminus S).}$$

A set ${\displaystyle U}$ is a dense open subset of ${\displaystyle X}$ if and only if ${\displaystyle \partial _{X}U=X\setminus U.}$

The interior of the boundary of a closed set is empty.[proof 1] 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.[proof 2] Consequently, the interior of the boundary of the interior of a set is empty. In particular, if ${\displaystyle S\subseteq X}$ is a closed or open subset of ${\displaystyle X}$ then there does not exist any nonempty subset ${\displaystyle U\subseteq \partial _{X}S}$ such that ${\displaystyle U}$ is open in ${\displaystyle X.}$ This fact is important for the definition and use of nowhere dense subsets, meager subsets, and Baire spaces.

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).

Boundary of hyperbolic components of Mandelbrot set

Consider the real line ${\displaystyle \mathbb {R} }$ with the usual topology (that is, the topology whose basis sets are open intervals) and ${\displaystyle \mathbb {Q} ,}$ the subset of rational numbers (whose topological interior in ${\displaystyle \mathbb {R} }$ is empty). Then

• ${\displaystyle \partial (0,5)=\partial [0,5)=\partial (0,5]=\partial [0,5]=\{0,5\}}$
• ${\displaystyle \partial \varnothing =\varnothing }$
• ${\displaystyle \partial \mathbb {Q} =\mathbb {R} }$
• ${\displaystyle \partial (\mathbb {Q} \cap [0,1])=[0,1]}$

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 ${\displaystyle \partial S}$ of a subset ${\displaystyle S}$ to contain a non-empty open subset of ${\displaystyle X:=\mathbb {R} }$; that is, for the interior of ${\displaystyle \partial S}$ in ${\displaystyle X}$ to be non-empty. However, a closed subset's boundary always has an empty interior.

In the space of rational numbers with the usual topology (the subspace topology of ${\displaystyle \mathbb {R} }$), the boundary of ${\displaystyle (-\infty ,a),}$ where ${\displaystyle a}$ is irrational, is empty.

The boundary of a set is a topological notion and may change if one changes the topology. For example, given the usual topology on ${\displaystyle \mathbb {R} ^{2},}$ the boundary of a closed disk ${\displaystyle \Omega =\left\{(x,y):x^{2}+y^{2}\leq 1\right\}}$ is the disk's surrounding circle: ${\displaystyle \partial \Omega =\left\{(x,y):x^{2}+y^{2}=1\right\}.}$ If the disk is viewed as a set in ${\displaystyle \mathbb {R} ^{3}}$ with its own usual topology, that is, ${\displaystyle \Omega =\left\{(x,y,0):x^{2}+y^{2}\leq 1\right\},}$ then the boundary of the disk is the disk itself: ${\displaystyle \partial \Omega =\Omega .}$ If the disk is viewed as its own topological space (with the subspace topology of ${\displaystyle \mathbb {R} ^{2}}$), then the boundary of the disk is empty.

This example demonstrates that the topological boundary of an open ball of radius ${\displaystyle r>0}$ is not necessarily equal to the corresponding sphere of radius ${\displaystyle r}$ (centered at the same point); it also shows that the closure of an open ball of radius ${\displaystyle r>0}$ is not necessarily equal to the closed ball of radius ${\displaystyle r}$ (again centered at the same point). Denote the usual Euclidean metric on ${\displaystyle \mathbb {R} ^{2}}$ by

$${\displaystyle d((a,b),(x,y)):={\sqrt {(x-a)^{2}+(y-b)^{2}}}}$$

which induces on ${\displaystyle \mathbb {R} ^{2}}$ the usual Euclidean topology.

Let ${\displaystyle X\subseteq \mathbb {R} ^{2}}$ denote the union of the ${\displaystyle y}$-axis ${\displaystyle Y:=\{0\}\times \mathbb {R} }$ with the unit circle

$${\displaystyle S^{1}:=\left\{p\in \mathbb {R} ^{2}:d(p,\mathbf {0} )=1\right\}=\left\{(x,y)\in \mathbb {R} ^{2}:x^{2}+y^{2}=1\right\}}$$

centered at the origin ${\displaystyle \mathbf {0}$ :=(0,0)\in \mathbb {R} ^{2}} ; that is, ${\displaystyle X:=Y\cup S^{1},}$ which is a topological subspace of ${\displaystyle \mathbb {R} ^{2}}$ whose topology is equal to that induced by the (restriction of) the metric ${\displaystyle d.}$

In particular, the sets ${\displaystyle Y,S^{1},Y\cap S^{1}=\{(0,\pm 1)\},}$ and ${\displaystyle \{0\}\times [-1,1]}$ are all closed subsets of ${\displaystyle \mathbb {R} ^{2}}$ and thus also closed subsets of its subspace ${\displaystyle X.}$ Henceforth, unless it clearly indicated otherwise, every open ball, closed ball, and sphere should be assumed to be centered at the origin ${\displaystyle \mathbf {0} =(0,0)}$ and moreover, only the metric space ${\displaystyle (X,d)}$ will be considered (and not its superspace ${\displaystyle (\mathbb {R} ^{2},d)}$); this being a path-connected and locally path-connected complete metric space.

Denote the open ball of radius ${\displaystyle r>0}$ in ${\displaystyle (X,d)}$ by ${\displaystyle B_{r}:=\left\{p\in X:d(p,\mathbf {0} )<r\right\}}$ so that when ${\displaystyle r=1}$ then

$${\displaystyle B_{1}=\{0\}\times (-1,1)}$$

is the open sub-interval of the ${\displaystyle y}$-axis strictly between ${\displaystyle y=-1}$ and ${\displaystyle y=1.}$ The unit sphere in ${\displaystyle (X,d)}$ ("unit" meaning that its radius is ${\displaystyle r=1}$) is

$${\displaystyle \left\{p\in X:d(p,\mathbf {0} )=1\right\}=S^{1}}$$

while the closed unit ball in ${\displaystyle (X,d)}$ is the union of the open unit ball and the unit sphere centered at this same point:

$${\displaystyle \left\{p\in X:d(p,\mathbf {0} )\leq 1\right\}=S^{1}\cup \left(\{0\}\times [-1,1]\right).}$$

However, the topological boundary ${\displaystyle \partial _{X}B_{1}}$ and topological closure ${\displaystyle \operatorname {cl} _{X}B_{1}}$ in ${\displaystyle X}$ of the open unit ball ${\displaystyle B_{1}}$ are:

$${\displaystyle \partial _{X}B_{1}=\{(0,1),(0,-1)\}\quad {\text{ and }}\quad \operatorname {cl} _{X}B_{1}~=~B_{1}\cup \partial _{X}B_{1}~=~B_{1}\cup \{(0,1),(0,-1)\}~=~\{0\}\times [-1,1].}$$

In particular, the open unit ball's topological boundary ${\displaystyle \partial _{X}B_{1}=\{(0,1),(0,-1)\}}$ is a proper subset of the unit sphere ${\displaystyle \left\{p\in X:d(p,\mathbf {0} )=1\right\}=S^{1}}$ in ${\displaystyle (X,d).}$ And the open unit ball's topological closure ${\displaystyle \operatorname {cl} _{X}B_{1}=B_{1}\cup \{(0,1),(0,-1)\}}$ is a proper subset of the closed unit ball ${\displaystyle \left\{p\in X:d(p,\mathbf {0} )\leq 1\right\}=S^{1}\cup \left(\{0\}\times [-1,1]\right)}$ in ${\displaystyle (X,d).}$ The point ${\displaystyle (1,0)\in X,}$ for instance, cannot belong to ${\displaystyle \operatorname {cl} _{X}B_{1}}$ because there does not exist a sequence in ${\displaystyle B_{1}=\{0\}\times (-1,1)}$ that converges to it; the same reasoning generalizes to also explain why no point in ${\displaystyle X}$ outside of the closed sub-interval ${\displaystyle \{0\}\times [-1,1]}$ belongs to ${\displaystyle \operatorname {cl} _{X}B_{1}.}$ Because the topological boundary of the set ${\displaystyle B_{1}}$ is always a subset of ${\displaystyle B_{1}}$'s closure, it follows that ${\displaystyle \partial _{X}B_{1}}$ must also be a subset of ${\displaystyle \{0\}\times [-1,1].}$

In any metric space ${\displaystyle (M,\rho ),}$ the topological boundary in ${\displaystyle M}$ of an open ball of radius ${\displaystyle r>0}$ centered at a point ${\displaystyle c\in M}$ is always a subset of the sphere of radius ${\displaystyle r}$ centered at that same point ${\displaystyle c}$; that is,

$${\displaystyle \partial _{M}\left(\left\{m\in M:\rho (m,c)<r\right\}\right)~\subseteq ~\left\{m\in M:\rho (m,c)=r\right\}}$$

always holds.

Moreover, the unit sphere in ${\displaystyle (X,d)}$ contains ${\displaystyle X\setminus Y=S^{1}\setminus \{(0,\pm 1)\},}$ which is an open subset of ${\displaystyle X.}$[proof 3] This shows, in particular, that the unit sphere ${\displaystyle \left\{p\in X:d(p,\mathbf {0} )=1\right\}}$ in ${\displaystyle (X,d)}$ contains a non-empty open subset of ${\displaystyle X.}$