Question

Let \((X,d)\) be a metric space. If \(S\) and \(T\) are subsets of \(X\), then which one of the options given below is NOT necessarily true? (For any \(Y\subseteq X\), \(\text{Int}(Y)\) denotes the interior of \(Y\) and \(\text{Clos}(Y)\) denotes the closure of \(Y\).)

• A. \(X\setminus \text{Int}(S) = \text{Clos}(X\setminus S)\)
• B. \(X\setminus \text{Clos}(S) = \text{Int}(X\setminus S)\)
• C. \(\text{Int}(S\cup T) = \text{Int}(S)\cup \text{Int}(T)\)
• D. \(\text{Clos}(S\cup T) = \text{Clos}(S)\cup \text{Clos}(T)\)

Hint:

The interior of a union can be strictly larger than the union of interiors; try two closed intervals that share only an endpoint.

Answer 1

The Correct Option is C

Step 1: Recall two standard duality identities for any subset \(Y\) of a metric space \(X\): \(\text{Int}(Y)=X\setminus \text{Clos}(X\setminus Y)\) and \(\text{Clos}(Y)=X\setminus \text{Int}(X\setminus Y)\). Rearranging the first gives \(X\setminus \text{Int}(S)=\text{Clos}(X\setminus S)\), which is option (A), always true. Rearranging the second gives \(X\setminus \text{Clos}(S)=\text{Int}(X\setminus S)\), which is option (B), always true.

Step 2: For closures of unions, \(\text{Clos}(S)\cup \text{Clos}(T)\) is a finite union of closed sets, hence closed, and it contains \(S\cup T\), so \(\text{Clos}(S\cup T)\subseteq \text{Clos}(S)\cup \text{Clos}(T)\). Conversely \(S\subseteq S\cup T\) gives \(\text{Clos}(S)\subseteq \text{Clos}(S\cup T)\), and similarly for \(T\), so \(\text{Clos}(S)\cup \text{Clos}(T)\subseteq \text{Clos}(S\cup T)\). Equality holds, so option (D) is always true.

Step 3: For interiors, \(\text{Int}(S)\) and \(\text{Int}(T)\) are open subsets of \(S\cup T\), so their union is an open subset of \(S\cup T\), giving only \(\text{Int}(S)\cup \text{Int}(T)\subseteq \text{Int}(S\cup T)\) in general, not equality.

Step 4: Counterexample in \(\mathbb{R}\) with the usual metric: let \(S=[0,1]\) and \(T=[1,2]\). Then \(S\cup T=[0,2]\), so \(\text{Int}(S\cup T)=(0,2)\), which contains the point \(1\). But \(\text{Int}(S)=(0,1)\) and \(\text{Int}(T)=(1,2)\), so \(\text{Int}(S)\cup \text{Int}(T)=(0,1)\cup(1,2)\), which excludes \(1\). Hence \(\text{Int}(S\cup T)\ne \text{Int}(S)\cup \text{Int}(T)\), so option (C) is NOT necessarily true.

\[\boxed{\text{Option (C) is not necessarily true}}\]