Equivalence of Definitions of Closure of Topological Subspace

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Theorem

The following definitions of the concept of Closure in the context of Topology are equivalent:


Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Definition 1

The closure of $H$ (in $T$) is defined as:

$H^- := H \cup H'$

where $H'$ is the derived set of $H$.

Definition 2

The closure of $H$ (in $T$) is defined as:

$\ds H^- := \bigcap \leftset {K \supseteq H: K}$ is closed in $\rightset T$

Definition 3

The closure of $H$ (in $T$), denoted $H^-$, is defined as the smallest closed set of $T$ that contains $H$.

Definition 4

The closure of $H$ (in $T$) is defined as the union of $H$ and its boundary in $T$:

$H^- := H \cup \partial H$

Definition 5

The closure of $H$ (in $T$) is the union of the set of all isolated points of $H$ and the set of all limit points of $H$:

$H^- := H^i \cup H'$

Definition 6

The closure of $H$ (in $T$), denoted $H^-$, is the set of all adherent points of $H$.


Proof

Definition $1$ is equivalent to Definition $2$

This is proved in Set Closure as Intersection of Closed Sets.

$\Box$


Definition $2$ is equivalent to Definition $3$

This is proved in Set Closure is Smallest Closed Set in Topological Space.

$\Box$


Definition $1$ is equivalent to Definition $4$

By the definition of the interior, and Set is Subset of its Topological Closure, it easily follows that

$H^\circ \subseteq H \subseteq H^-$

Then:

\(\ds H \cup \partial H\) \(=\) \(\ds H \cup \paren {H^- \setminus H^\circ}\) Definition of Boundary (Topology)
\(\ds \) \(=\) \(\ds H \cup \paren {H \setminus H^\circ} \cup \paren {H^- \setminus H}\) Union of Relative Complements of Nested Subsets
\(\ds \) \(=\) \(\ds H \cup \paren {H^- \setminus H}\) Set Difference is Subset and Union with Superset is Superset
\(\ds \) \(=\) \(\ds H^-\) Union with Relative Complement

$\Box$


Definition $1$ is equivalent to Definition $5$

Every isolated point of $H$ is a point of $H$.

So by Set Union Preserves Subsets:

$(1): \quad H^i \cup H' \subseteq H \cup H'$


If $S \setminus \paren {H^i \cup H'} = \O$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.


Otherwise, let $x \in S \setminus \paren {H^i \cup H'} \ne \O$.

From De Morgan's Laws:

$S \setminus \paren {H^i \cup H'} = \paren {S \setminus H^i} \cap \paren {S \setminus H'}$

Thus $x$ is a point of $S$ that is neither an isolated point of $H$ nor a limit point of $H$.


Then there exists an open set $U$ that contains $x$ such that:

$U \cap H \ne \set x$

and:

$H \cap \paren {U \setminus \set x} = \O$


This implies that

$U \cap H = \O$

By Condition for Point being in Closure:

$x \notin H \cup H'$


It has been shown that:

$x \notin H^i \cup H' \implies x \notin H \cup H'$

From Rule of Transposition:

$x \in H \cup H' \implies x \in H^i \cup H'$

And so

$(2): \quad H \cup H' \subseteq H^i \cup H'$


Combining $(1)$ and $(2)$:

$H \cup H' = H^i \cup H'$

Hence the result.

$\Box$


Definition $1$ is equivalent to Definition $6$

By one of the equivalent definitions of an adherent point:

A point $x \in S$ is an adherent point of $H$ if and only if every open neighborhood $U$ of $x$ satisfies $H \cap U \ne \O$


From Condition for Point being in Closure:

$x \in H^-$ if and only if every open set of $T$ which contains $x$ contains a point in $H$

where:

$H^-$ is the union of $H$ and all the limit points of $H$ in $T$

The equivalence follows.

$\blacksquare$


Sources

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