Let $X$ be a topological space, and let $S$ be a subset of $X$. Then $\mathrm{int} S$, by definition, is the union of all the open sets of $X$ that are contained in $S$.

Now let $A$ and $B$ be any subsets of $X$. Then what is the relation, if any, between (i) $\mathrm{Int} (A \cup B)$ and $\mathrm{Int} A \cup \mathrm{Int} B$? (ii) $\mathrm{Int} (A \cap B)$ and $\mathrm{Int} A \cap \mathrm{Int} B$? (iii) $\mathrm{Int} (A \setminus B)$ and $\mathrm{Int} A \setminus \mathrm{Int} B$? (iv) $\mathrm{Int} (A \Delta B)$ and $\mathrm{Int} A \Delta \mathrm{Int} B$? The set $A \Delta B$ is defined as follows: $$ A \Delta B \colon= ( A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (A \cap B). $$

My Attempt:

We first show the following: If $S$ and $T$ are any subsets of $X$ such that $S \subset T$, then $\mathrm{Int} S \subset \mathrm{Int} T$.

So suppose that $S \subset T \subset X$. Let $x$ be any interior point of $S$. Then there exists an open set $U_x$ in $X$ such that $x \in U_x \subset S$. But $S \subset T$. So we can conclude that $U_x \subset T$, which implies that this $x$ is an interior point of $T$ also.

We also note that $\mathrm{Int} \emptyset = \emptyset$, because the only open set contained in $\emptyset$ is itself.

(i) Now as $A \subset A \cup B$ and $B \subset A \cup B$, so we must have $\mathrm{Int} A \subset \mathrm{Int} (A \cup B)$ and $\mathrm{Int} B \subset \mathrm{Int} (A \cup B)$, and hence $$\mathrm{Int} A \cup \mathrm{Int} B \subset \mathrm{Int} (A \cup B). \tag{0} $$ The reverse inclusion may not hold. For example, let $X = \mathbb{R}$, with the standard topology; let $A = \mathbb{Q}$ and $B = \mathbb{R} \setminus \mathbb{Q}$. Then $A \cup B = \mathbb{R}$ so that $\mathrm{Int} (A \cup B) = \mathbb{R}$, whereas $\mathrm{Int} A = \emptyset = \mathrm{Int} B$ and so $\mathrm{Int} A \cup \mathrm{Int} B = \emptyset$.

Under what most general possible condition(s) on $X$, if any, does the reverse inclusion also hold for every pair $A$, $B$ of subsets of $X$?

(ii) As $A \cap B \subset A$ and $A \cap B \subset B$, so we must have $\mathrm{Int} (A \cap B) \subset \mathrm{Int} A$ and $\mathrm{Int} (A \cap B) \subset \mathrm{Int} B$, which implies that $$\mathrm{Int} (A \cap B) \subset \mathrm{Int} A \cap \mathrm{Int} B. \tag{1} $$ Now let $x \in \mathrm{Int} A \cap \mathrm{Int} B$. Then $x \in \mathrm{Int} A$ and $x \in \mathrm{Int} B$, which implies that there are open sets $U_x$ and $V_x$ in $X$ such that $x \in U_x \subset A$ and $x \in V_x \subset B$; therefore $U_x \cap V_x$ is an open set in $X$ and $x \in U_x \cap V_x \subset A \cap B$, which implies that $x \in \mathrm{Int} (A \cap B)$. Thus we have shown that $$ \mathrm{Int} A \cap \mathrm{Int} B \subset \mathrm{Int} (A \cap B). \tag{2} $$ From (1) and (2) we can conclude that $$ \mathrm{Int} (A \cap B) = \mathrm{Int} A \cap \mathrm{Int} B. \tag{A} $$

Am I right?

(iii) As $A \setminus B \subset A$, so $\mathrm{Int} (A \setminus B) \subset \mathrm{Int} A$. Now let $x \in \mathrm{Int} (A \setminus B)$. Then $x \in \mathrm{Int} A$. We now show that this $x$ cannot be in $\mathrm{Int} B$. So suppose that $x \in \mathrm{Int} B$. Then by (A) above we can conclude that $$ x \in \mathrm{Int} A \cap \mathrm{Int} B = \mathrm{Int} (A \cap B). $$ But as $x \in \mathrm{Int} (A \setminus B)$, by our choice of $x$, so again by (A) above we can conclude that $$ x \in \mathrm{Int} (A \cap B) \cap \mathrm{Int} (A \setminus B) = \mathrm{Int} \left( (A \cap B) \cap (A \setminus B) \right) = \mathrm{Int} (\emptyset),$$ which is a contradiction, because $\mathrm{Int} (\emptyset) = \emptyset$. Thus our supposition that $x \in \mathrm{Int} (A \setminus B)$ and $x \in \mathrm{Int} B$ is wrong. So if $x \in \mathrm{Int} (A \setminus B)$, then $x \in \mathrm{Int} A$ but $x \not\in \mathrm{Int} B$, which implies that $x \in \mathrm{Int} A \setminus \mathrm{Int} B$. Therefore, we have $$ \mathrm{Int} (A \setminus B) \subset \mathrm{Int} A \setminus \mathrm{Int} B. \tag{3} $$ The reverse inclusion may not hold. For example, let $X = \mathbb{R}$, with the standard topology; let $A = [0, 2]$ and $B = [1, 3]$; then $\mathrm{Int} A = (0, 2)$ and $\mathrm{Int} B = (1, 3)$ so that $$\mathrm{Int} A \setminus \mathrm{Int} B = (0, 1]; $$ on the other hand, we note that $A \setminus B = [0, 1)$, and so $$ \mathrm{Int} (A \setminus B) = (0, 1) \subsetneqq (0, 1] = \mathrm{Int} A \setminus \mathrm{Int} B. $$

Am I right? Under what most general condition(s) on $X$, if any, does the reverse inclusion also hold for every pair of subsets $A$ and $B$ of $X$?

(iv) We note that $$ \begin{align} \mathrm{Int} A \Delta \mathrm{Int} B &= \left( \mathrm{Int} A \setminus \mathrm{Int} B \right) \cup \left( \mathrm{Int} B \setminus \mathrm{Int} A \right) \\ &\subset \mathrm{Int} \left( ( A \setminus B) \cup (B \setminus A) \right) \qquad \mbox{ [ using (0) above ] } \\ &= \mathrm{Int} ( A \Delta B). \tag{4} \end{align} $$ The reverse inclusion may not hold, however. As an example, take the same topological space $X$ and the same sets $A$ and $B$ as in (iii) above. We note that $A \Delta B = [0, 1) \cup (2, 3]$ so that $\mathrm{Int} (A \Delta B) = (0, 1) \cup (2, 3)$; on the other hand, $$ \mathrm{Int} A \Delta \mathrm{Int} B = (0, 2) \Delta (1, 3) = (0, 1] \cup [2, 3) \supsetneqq (0, 1) \cup (2, 3) = \mathrm{Int} (A \Delta B). $$

Am I right? What is the most general condition(s) on $X$ for the reverse inclusion to hold for every pair $A$ and $B$ of subsets of $X$?

  • $\begingroup$ Why a close-vote, I wonder? $\endgroup$ – Saaqib Mahmood Dec 13 '17 at 21:18
  • 1
    $\begingroup$ point 4 honestly i don't get the proof, also in the example you are showing a counterexample of what you have just asserted. Maybe i'm missing something but you say that $IntA \Delta IntB\subset Int(A\Delta B)$ and then in the example show that $Int(A\Delta B)) \subsetneq IntA \Delta IntB$ $\endgroup$ – chak Dec 13 '17 at 21:33
  • $\begingroup$ Proof verifications are, the last time I checked, perfectly acceptable. Most of these proofs can be simplified if you understand that the interior of the set is "the largest open set contained in that set". Thus $Int(A)\cup Int(B)\subseteq Int(A\cup B)$ is clear because the former is an open set contained in $A\cup B$ thus must be contained in the largest open set contained in $A\cup B$, which is $Int(A\cup B)$. The same reasoning makes all of part $(ii)$ clear. I'd take a closer look at (iv). How exactly did you derive that first inclusion (interiors don't commute with differences)? $\endgroup$ – user123641 Dec 13 '17 at 22:50
  • $\begingroup$ @Robert thank you for your comment. There is a mistake in my attempt at (iv). I'll correct it later. $\endgroup$ – Saaqib Mahmood Dec 14 '17 at 6:34
  • $\begingroup$ @chak I've made a mistake in (iv), which I'll rectify in a while. $\endgroup$ – Saaqib Mahmood Dec 14 '17 at 6:35

This is what i came up with.

For the union i think i have a condition but i don't know (and i don't think) if it is general.

Take $A,B$ subsets of $X$, then if $\exists U,V$ open sets such that $A \subsetneq U, B\setminus A\subsetneq V, U \cap V=\emptyset$ then we have that $IntA \cup IntB= Int(A \cup B) $.

We know that one side is true, now if $IntA \cup IntB \subsetneq Int(A \cup B)$ there must be an open set $K$ that is contained in $Int(A \cup B)$ but it is not contained in $IntA \cup IntB$.

Now we have that $K \cap U \subset A$ is an open set and $K \cap V \subset B$ it is an open set as well, we also have that $K=(K \cap U) \cup (K \cap V)$ since $K \subset A \cup B \subset U \cup V$, so $K$ is the union of two open sets respectively in $IntA$ and in $IntB$ and so $K $ is in $Int(A \cup B)$. We have then that $IntA \cup IntB= Int(A \cup B).$

$Int(A \setminus B)=Int(A \cap B^c)=Int(A) \cap Int(B^c)$, $Int(A) \setminus Int(B)=Int(A) \cap (Int(B))^c$ so a sufficient condition would be $Int(B^c)=(IntB)^c$, obviously it is not necessary.


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