Editing Boundary (topology) (section)

== Examples ==

=== Characterizations and general examples ===

A set and its complement have the same boundary:
<math display="block">\partial_X S = \partial_X (X \setminus S).</math>

A set <math>U</math> is a [[Dense subset|dense]] [[Open set|open]] subset of <math>X</math> if and only if <math>\partial_X U = X \setminus U.</math> 

The interior of the boundary of a closed set is empty.<ref group="proof">Let <math>S</math> be a closed subset of <math>X</math> so that <math>\overline{S} = S</math> and thus also <math>\partial_X S := \overline{S} \setminus \operatorname{int}_X S = S \setminus \operatorname{int}_X S.</math> If <math>U</math> is an open subset of <math>X</math> such that <math>U \subseteq \partial_X S</math> then  <math>U \subseteq S</math> (because <math>\partial_X S \subseteq S</math>) so that <math>U \subseteq \operatorname{int}_X S</math> (because [[Interior (topology)|by definition]], <math>\operatorname{int}_X S</math> is the largest open subset of <math>X</math> contained in <math>S</math>). But <math>U \subseteq \partial_X S = S \setminus \operatorname{int}_X S</math> implies that <math>U \cap \operatorname{int}_X S = \varnothing.</math> Thus <math>U</math> is simultaneously a subset of <math>\operatorname{int}_X S</math> and disjoint from <math>\operatorname{int}_X S,</math> which is only possible if <math>U = \varnothing.</math> [[Q.E.D.]]</ref> 
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.<ref group="proof">Let <math>S</math> be an open subset of <math>X</math> so that <math>\partial_X S := \overline{S} \setminus \operatorname{int}_X S = \overline{S} \setminus S.</math> Let <math>U := \operatorname{int}_X \left(\partial_X S\right)</math> so that <math>U = \operatorname{int}_X \left(\partial_X S\right) \subseteq \partial_X S = \overline{S} \setminus S,</math> which implies that <math>U \cap S = \varnothing.</math> If <math>U \neq \varnothing</math> then pick <math>u \in U,</math> so that <math>u \in U \subseteq \partial_X S \subseteq \overline{S}.</math> Because <math>U</math> is an open neighborhood of <math>u</math> in <math>X</math> and <math>u \in \overline{S},</math> the definition of the [[Closure (topology)|topological closure]] <math>\overline{S}</math> implies that <math>U \cap S \neq \varnothing,</math> which is a contradiction. <math>\blacksquare</math> Alternatively, if <math>S</math> is open in <math>X</math> then <math>X \setminus S</math> is closed in <math>X,</math> so that by using the general formula <math>\partial_X S = \partial_X (X \setminus S)</math> and the fact that the interior of the boundary of a closed set (such as <math>X \setminus S</math>) is empty, it follows that <math>\operatorname{int}_X \partial_X S = \operatorname{int}_X \partial_X (X \setminus S) = \varnothing.</math> <math>\blacksquare</math></ref> 
Consequently, the interior of the boundary of the interior of a set is empty. 
In particular, if <math>S \subseteq X</math> is a closed or open subset of <math>X</math> then there does not exist any nonempty subset <math>U \subseteq \partial_X S</math> such that <math>U</math> is open in <math>X.</math> 
This fact is important for the definition and use of [[Nowhere dense set|nowhere dense subsets]], [[Meager set|meager subsets]], and [[Baire space]]s. 

A set is the boundary of some open set if and only if it is closed and [[Nowhere dense set|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]]).

=== Concrete examples ===

[[File:Mandelbrot Components.svg|right|thumb|Boundary of hyperbolic components of [[Mandelbrot set]]]]
Consider the real line <math>\R</math> with the usual topology (that is, the topology whose [[Basis (topology)|basis sets]] are [[open interval]]s) and <math>\Q,</math> the subset of rational numbers (whose [[Interior (topology)|topological interior]] in <math>\R</math> is empty).  Then

* <math>\partial (0,5) = \partial [0,5) = \partial (0,5] = \partial [0,5] = \{0, 5\}</math>
* <math>\partial \varnothing= \varnothing</math>
* <math>\partial \Q = \R</math>
* <math>\partial (\Q \cap [0, 1]) = [0, 1]</math>

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

In the space of rational numbers with the usual topology (the [[subspace topology]] of <math>\R</math>), the boundary of <math>(-\infty, a),</math> where <math>a</math> is irrational, is empty.

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

=== Boundary of an open ball vs. its surrounding sphere ===

This example demonstrates that the topological boundary of an open ball of radius <math>r > 0</math> is {{em|not}} necessarily equal to the corresponding sphere of radius <math>r</math> (centered at the same point); it also shows that the closure of an open ball of radius <math>r > 0</math> is {{em|not}} necessarily equal to the closed ball of radius <math>r</math> (again centered at the same point). 
Denote the usual [[Euclidean metric]] on <math>\R^2</math> by
<math display="block">d((a, b), (x, y)) := \sqrt{(x - a)^2 + (y - b)^2}</math>
which induces on <math>\R^2</math> the usual [[Euclidean topology]]. 
Let <math>X \subseteq \R^2</math> denote the union of the <math>y</math>-axis <math>Y := \{ 0 \} \times \R</math> with the unit circle <math display="block">S^1 := \left\{ p \in \R^2 : d(p, \mathbf{0}) = 1 \right\} = \left\{ (x, y) \in \R^2 : x^2 + y^2 = 1 \right\}</math> centered at the origin <math>\mathbf{0} := (0, 0) \in \R^2</math>; that is, <math>X := Y \cup S^1,</math> which is a [[topological subspace]] of <math>\R^2</math> whose topology is equal to that induced by the (restriction of) the metric <math>d.</math> 
In particular, the sets <math>Y, S^1, Y \cap S^1 = \{ (0, \pm 1) \},</math> and <math>\{ 0 \} \times [-1, 1]</math> are all closed subsets of <math>\R^2</math> and thus also closed subsets of its subspace <math>X.</math> 
Henceforth, unless it clearly indicated otherwise, every open ball, closed ball, and sphere should be assumed to be centered at the origin <math>\mathbf{0} = (0, 0)</math> and moreover, only the [[metric space]] <math>(X, d)</math> will be considered (and not its superspace <math>(\R^2, d)</math>); this being a [[Path-connected space|path-connected]] and [[locally path-connected]] [[complete metric space]]. 

Denote the open ball of radius <math>r > 0</math> in <math>(X, d)</math> by 
<math>B_r := \left\{ p \in X : d(p, \mathbf{0}) < r \right\}</math>
so that when <math>r = 1</math> then 
<math display="block">B_1 = \{ 0 \} \times (-1, 1)</math>
is the open sub-interval of the <math>y</math>-axis strictly between <math>y = -1</math> and <math>y = 1.</math> 
The unit sphere in <math>(X, d)</math> ("unit" meaning that its radius is <math>r = 1</math>) is 
<math display="block">\left\{ p \in X : d(p, \mathbf{0}) = 1 \right\} = S^1</math>
while the closed unit ball in <math>(X, d)</math> is the union of the open unit ball and the unit sphere centered at this same point: 
<math display="block">\left\{ p \in X : d(p, \mathbf{0}) \leq 1 \right\} = S^1 \cup \left(\{ 0 \} \times [-1, 1]\right).</math>

However, the topological boundary <math>\partial_X B_1</math> and topological closure <math>\operatorname{cl}_X B_1</math> in <math>X</math> of the open unit ball <math>B_1</math> are:
<math display="block">\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].</math>
In particular, the open unit ball's topological boundary <math>\partial_X B_1 = \{ (0, 1), (0, -1) \}</math> is a {{em|proper}} subset of the unit sphere <math>\left\{ p \in X : d(p, \mathbf{0}) = 1 \right\} = S^1</math> in <math>(X, d).</math> 
And the open unit ball's topological closure <math>\operatorname{cl}_X B_1 = B_1 \cup \{ (0, 1), (0, -1) \}</math> is a proper subset of the closed unit ball <math>\left\{ p \in X : d(p, \mathbf{0}) \leq 1 \right\} = S^1 \cup \left(\{ 0 \} \times [-1, 1]\right)</math> in <math>(X, d).</math> 
The point <math>(1, 0) \in X,</math> for instance, cannot belong to <math>\operatorname{cl}_X B_1</math> because there does not exist a sequence in <math>B_1 = \{ 0 \} \times (-1, 1)</math> that converges to it; the same reasoning generalizes to also explain why no point in <math>X</math> outside of the closed sub-interval <math>\{ 0 \} \times [-1, 1]</math> belongs to <math>\operatorname{cl}_X B_1.</math> Because the topological boundary of the set <math>B_1</math> is always a subset of <math>B_1</math>'s closure, it follows that <math>\partial_X B_1</math> must also be a subset of <math>\{ 0 \} \times [-1, 1].</math> 

In any metric space <math>(M, \rho),</math> the topological boundary in <math>M</math> of an open ball of radius <math>r > 0</math> centered at a point <math>c \in M</math> is always a subset of the sphere of radius <math>r</math> centered at that same point <math>c</math>; that is, 
<math display="block">\partial_M \left(\left\{ m \in M : \rho(m, c) < r \right\}\right) ~\subseteq~ \left\{ m \in M : \rho(m, c)= r \right\}</math>
always holds. 

Moreover, the unit sphere in <math>(X, d)</math> contains <math>X \setminus Y = S^1 \setminus \{ (0, \pm 1) \},</math> which is an open subset of <math>X.</math><ref group="proof">The <math>y</math>-axis <math>Y = \{ 0 \} \times \R</math> is closed in <math>\R^2</math> because it is a product of two closed subsets of <math>\R.</math> Consequently, <math>\R^2 \setminus Y</math> is an open subset of <math>\R^2.</math> Because <math>X</math> has the subspace topology induced by <math>\R^2,</math> the intersection <math>X \cap \left(\R^2 \setminus Y\right) = X \setminus Y</math> is an open subset of <math>X.</math> <math>\blacksquare</math></ref> This shows, in particular, that the unit sphere <math>\left\{ p \in X : d(p, \mathbf{0}) = 1 \right\}</math> in <math>(X, d)</math> contains a {{em|non-empty open}} subset of <math>X.</math>