The closure of a subspace

Let $X$ be a topological space. If $A$ is a subspace of $X$ we denote its closure by $\overline A$. For each point $x \in X$ the family $N_x$ of neighbourhoods of $x$ is a filter on $X$, the $\textit{neighbourhood filter}$ of $x$.

This is from Bell and Slomson (1969) Models and Ultraproducts: an introduction, p.22, who prove $\textit{inter alia}$ that, where $F$ is a filter on $X$, (i) and (ii) are equivalent:

(i) $x \in \bigcap\{\overline A: A \in F\}$

(ii) For all $A \in F$ and $U \in N_x, A \cap U \neq \emptyset$

By $\overline A$ do they mean the intersection of all the closed sets of the subspace topology $(A, \tau_A$) that contain $A$; or, alternatively, do they simply mean the set of closed sets of the subspace topology? That is, suppose we have a topological space $(X, \tau)$ with $X = \{a, b, c, d, e\}$ and topology $\tau = \{X, \emptyset, \{a\}, \{d\}, \{a, d\} \}$, and suppose we have a subspace topology $(A, \tau_A$) of the topological space $(X, \tau)$, with $A= \{a, c\}$ and $\tau_A = \{A, \emptyset,\{a\} \}$. The closed sets of $(A, \tau_A$) are then:

$A - A = \emptyset, \hspace{0.8cm} A - \emptyset = A, \hspace{0.8cm} A - \{a\} = \{c\}$

The intersection of all these closed sets containing $A$ is then $A$. So the closure by $\overline A$ in this case would be $A$. Is that right?

Or, alternatively, by the "closure by $\overline A$" do Bell and Slomson just mean the set of closed sets of $(A, \tau_A$), with $A= \{a, c\}$; i.e, the set containing:

$A - A = \emptyset, \hspace{0.8cm} A - \emptyset = A, \hspace{0.8cm} A - \{a\} = \{c\}$

When you write $\overline{A}$ in your last sentence, do you mean the closure in the subspace topology? This would be A, as you said and your argumentation is correct. Or the closure in the topology of X? In this case: $\overline{A} = \{a,b,c,e\}$.

• That's my question: how are Bell and Slomson using $\overline$ in $\overline A$? – user378426 Nov 27 '17 at 10:55
• I think they are using it as closure in the topology of $X$. The closure of a subset in it's own induced subspace topology is always the subset itself. That would not make any sense in the above formulated equivalence. – Falrach Nov 27 '17 at 10:56
• Sorry about this, but what do you mean by the closure in the topology of $X$? – user378426 Nov 27 '17 at 10:59
• Does that mean the intersection of all the closed sets of $(X, \tau)$ that contain $A\hspace{0.2cm}$(i.e, that contain $\{ a, c \}$? – user378426 Nov 27 '17 at 11:01
• Yes, that's it. For a subset $B \subset A$ we could write: $\overline{B}^X = \cap_{C \subset X closed : B \subset C} C$ for the closure of $B$ in the topology of $X$ and $\overline{B}^A= \cap_{C \subset X closed : B \subset C} C$ for the closure of $B$ in the induced topology of $A$. In general $\overline{B}^X$ does not have to be in $A$ anymore. – Falrach Nov 27 '17 at 11:02

$\overline A$ means the closure of the subset A within the space X.
"A is a subspace of X" should read "A is a subset of X".

When viewing A as a space the expression "A is a closed subspace" means the subset A is closed within the larger space.

By $\overline{A}$ they mean the closure of the subset $A$ in the space $X$.

This set $\overline{A}$ can be defined in two equivalent ways:

1. $\overline{A} = \bigcap\{C \subseteq X: C \text{ closed in } X \text { and } A \subseteq C\}$

2. $\overline{A} = \{x \in X: \forall U \in N_x: U \cap A \neq \emptyset\}$

$X = \{a, b, c, d, e\}$ and topology $\tau = \{X, \emptyset, \{a\}, \{d\}, \{a, d\} \}$ and $A = \{a,c\}$.

Compute the closed sets (the complements of the open sets), these are

$$\emptyset, X, \{b,c,d,e\}, \{a,b,c,e\}, \{b,c,e\}$$

and the ones that contain $A$ are just: $X$, $\{a,b,c,e\}$ and so $\overline{A} = \{a,b,c,e\}$, by applying formula $1$.

Now (in general) if $x \in A$, every member of $N_x$ contains $x$ and thus intersects $A$ trivially. So $a \in \overline{A}, c \in \overline{A}$ according to definition $2$. We also see $N_b = \{X\}$ and so trivially, every member of it intersects $A$ and so $b \in \overline{A}$. For $d$ we have $N_d = \{A \subseteq X: d \in A\}$ (the filter generated by $\{d\}$) and so $X\setminus A \in N_d$ and is disjoint from $A$, so by definition $2$, $d \notin \overline{A}$. Finally $N_e = \{X\}$ again, so as before $e \in \overline{A}$. So also definition $2$ gives us $\overline{A} = \{a,b,c,e\}$.

This holds in general (that these defintions yield the same sets), and is classical. It should IMHO be part of any topology course.

From definition $2$. the proof in your book is almost immediate.

It must be that $\bar A$ denotes $Cl_X(A)$... The closure of $A$ in the subspace $A$ is just $A$ itself. If, in (i), we replace $\bar A$ with $A$ (...thinking that $\bar A$ means $Cl_A(A),$ which is $A$... ) then (i) says $x\in \cap F.$ But if we do that then the result is false.

For example let $X=\Bbb R$ with the usual topology, let $x=0,$ and let $S\subset \Bbb R$ belong to $F$ iff $\exists r>0\;(S\supset [-r,0)\cup (0,r]).$ Then (ii) holds but $\cap F=\phi.$