Containment of sets in arbitrary metric spaces

Question

This is the full question:

$(X,d)$ be any arbitrary metric space. Let $S,T \subseteq X$. Prove or disprove the following:(if it's incorrect, what will be the correct statement?)

$1)S^o\cap T^o = (S\cap T)^o\,\,$ $\,\,2)S^o \cup T^o = (S\cup T)^o $

$3)$ $\overline{S\cap T}$ $\subseteq$ $\overline{S}\cap\overline{T}\,\,$ $\,\,4) \overline{S\cup T}$ = $\overline{S}\cup\overline{T}\,\,$ $\,\,5)$ $(S^o)^c = \overline{(S^c)}$

$6)$ $(S)^o = \left(\overline{S}\right)^o\,\,$ $\,\,7)$ $\overline{S} = \overline{\left(S^o\right)}\,\,$ $\,\,8)$ $\overline{S- T}$ = $\overline{S}-\overline{T}$

$9)$ If $(X,||.||)$ is a Norm Linear space then is it true that $\overline{S+ T}$ = $\overline{S}+\overline{T}$ ?

$1)$ is true. $S^o \subseteq S $ and $T^o \subseteq T $. This implies $S^o \cap T^o \subseteq S \cap T$, and since the interior of a set is the largest open set contained in it, we have $S^o \cap T^o \subseteq \left( S \cap T \right)^o$ $\subseteq S \cap T$. Also $S \cap T \subseteq S$ and $S \cap T \subseteq T$. This implies $\left(S \cap T\right)^o \subseteq S^o$ and $\left(S \cap T\right)^o \subseteq T^o$. Hence $\left(S \cap T\right)^o \subseteq S^o \cap T^o$. This finishes the proof of $1)$.

$2)$ is false, a counterexample will be $S=\mathbb{Q}$ and $T=\mathbb{Q^c}$ in $X=\mathbb{R}$. $\,S^o=\emptyset$, $T^o=\emptyset$ and $S \cup T =$ $\mathbb{R}$. It should be $S^o \cup T^o \subseteq (S\cup T)^o$. $S^o \subseteq S $ and $T^o \subseteq T $, so $S^o \cup T^o$ $\subseteq$ $S \cup T$. Which implies $\left( S^o \cup T^o \right)^o$ $=$ $ S^o \cup T^o$ $\subseteq$ $\left( S \cup T\right)^o$.

$3)$ is true. $S\cap T$ $\subseteq S$ and $S\cap T$ $\subseteq$ $T$. Which implies $\overline{S\cap T}$ $\subseteq$ $\overline{S}$ and $\overline{S\cap T}$ $\subseteq$ $\overline{T}$. Hence $\overline{S\cap T}$ $\subseteq$ $\overline{S}\cap\overline{T}$. This finishes proof of $3)$. It is not necessary that $\overline{S\cap T}$ $=$ $\overline{S}\cap\overline{T}$, counter example of $2)$ will work.

$4)$ I am not sure. $S$ $\subseteq S\cup T$ and $T$ $\subseteq S\cup T$. which implies $\overline{S}$ $\subseteq \overline{S\cup T}$ and $\overline{T}$ $\subseteq \overline{S\cup T}$, and hence $\overline{S}\cup\overline{T}$ $\subseteq$ $\overline{S\cup T}$. It seems to be false, but i am not able to find a counter example.

$6)$ and $7)$ are false. A counterexample would be $S =\mathbb{Q}$ in $X=\mathbb{R}$. I think the correct containments are $(S)^o$ $\subseteq$ $\left(\overline{S}\right)^o$ and $\overline{S}$ $\supseteq$ $\overline{\left(S^o\right)}$. Proof: $S \subset \overline{S}$ , which implies $S^o \subseteq \left(\overline{S}\right)^o$ $\subseteq$ $\overline{S}$. Also, $S^o \subseteq S$ hence $\overline{\left(S^o\right)}$ $\subseteq$ $\overline{S}$.

$8)$ is false. A counterexample will be $S=[0,1]$ and $T=[0,1)$ in $X=\mathbb{R}$. Will the correct containment be $\overline{S- T}$ $\supseteq$ $\overline{S}-\overline{T}$ ?

I am stuck with $5)$ and $9)$. Kindly check whether the above proofs and counterexamples are correct or not. Any alternate easy proofs will be great :).

Definition 1: A point $x \in X$ is called a limit point of $E\subseteq X$ if $B(x,r)\cap E \ne \emptyset $ $\,\forall\, r>0$.

Definition 2:(Closure of a set) Let $(X, d)$ be a metric space and $A\subset X$. Let $\overline{A}$ denote the set of all limit points of $A$. Thus $\overline{A}$ $= \{y \in X : $ For every $r > 0$, the set $B(y, r) \cap A \ne\emptyset \}$.

Definition 3(Interior of a set): Let $S\subset X$ be a subset of a metric space. We say that $x \in S$ is an interior point of $S$ if $\,\exists\,$ $r > 0$ such that $B(x, r) \subset S$ . The set of interior points of $S$ is denoted by $S^o$ and is called the interior of the set $S$

What you have so far is correct. Except for your hunch in 4); 4) is equivalent to 1), by taking complements [that is 5), which you don not yet have]. — Daniel Fischer, Jan 24 '20 at 14:18
4,5 are true, the proofs are not hard. $\overline{S} + \overline{T} \subseteq \overline{S+T}$ follows from continuity of addition. The other inclusion can fail in general, IIRC. — Henno Brandsma, Jan 24 '20 at 23:42

score 1 · Answer 1 · answered Jan 24 '20 at 15:35

Number 5 is an important equation for all topological spaces.
Proof.
x in X - S$^o$ iff x not in S$^o$ iff
not exists open U nhood x with U subset S iff
for all open U nhood x, not-(U subset S) iff
for all open U nhood x, exists y in U with y not in S iff
for all open U nhood x, exists y in U $\cap$ (X - S) iff
for all open U nhood x, U $\cap$ (X - S) is not empty iff
x in $\overline {X - S}$.

For metric spaces instead of open neighborhoods
of x, one may use open balls centered at x.

The usual definition of a limit points is:
x is a limit point of A when
for all open U nhood A, exists y in U $\cap$ A with y /= x.
What you called a limit point is usually called an adherence point.

score 1 · Accepted Answer · answered Jan 24 '20 at 23:50

Suppose $x \notin \overline{S}$ and $x \notin \overline{T}$. There then exist $r,s>0$ such that $B(x,r) \cap S= \emptyset$ and $B(x,s) \cap T = \emptyset$, but then $B(x, \min(r,s)) \cap (S \cup T) = \emptyset$ too and so $x \notin \overline{S \cup T}$. This shows $\overline{S \cup T} \subseteq \overline{S} \cup \overline{T}$ by contradiction.

nr 5. follows from the observation that if no ball around $x$ stays inside $S$ (so $x \notin S^\circ$) then every ball around $x$ intersects $S^c$, so $x \in \overline{S^c}$ and vice versa.

Containment of sets in arbitrary metric spaces

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