Let $A = [0,1) \cup (3,4]$ be a subset of $(\mathbb R, \mathfrak T_H)$

Let $A = [0,1) \cup (3,4]$ be a subset of $(\mathbb R, \mathfrak T_H)$

$\mathfrak T_H$ is the collection of all subsets of $U$ of $\mathbb R$ such that either $U = \emptyset$ or for each $x \in U$ there is an interval of the form $[a,b)$ such that $x \in [a, b) \subseteq U$.

Find

$Int(A)= [0,1)$

$Cl(A)=$

$Ext(A)= (-\infty, 0)\cup (1,3) \cup (4, \infty)$

$Bd(A)=$

I have a really hard time picking out these sets in anything other than the usual topology.

My defintions are as follows: Interior- Let $(X, \mathfrak T)$ be a topological space and let $A \subset X$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.

Closure- Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$ . The closure of $A$ is $Cl(A) = \bigcap \{U \subseteq X: U$ is a closed set and $A \subseteq U\}$ Based on this I know $A \subseteq Cl(A)$

Exterior- Boundary- $A'$ is the set of all limit points and my definition for this is: Let $(X, \mathfrak T)$ be a topological space with $A \subseteq X$. A point $x$ in $X$ is said to be a limit point of $A$ provided that every open set containing $x$ contains a point $A$ different from $x$.

• Do all intervals of the sets (closure, interior, exterior) have to be in the form $[a,b)$ or $\emptyset$? – user219081 May 15 '15 at 20:20
• The closure is the smallest closed set containing A. The set A is not closed in the half-open interval topology as only intervals of the form $[a,b)$ closed – user219081 May 15 '15 at 20:57
• I believe sets of the form $[a,b)$ are open, as trivially for every $x \in [a,b)$, $[a,b)$ is a set containing $x$ with $[a,b) \subset [a,b)$. I am also suspecting that this topology is equivalent to the lower limit topology. If so, we are in luck, because it is a refinement of the standard topology on $\mathbb{R}$, which is easier to conceptualize. – graydad May 15 '15 at 20:57
• My book says sets in the form $[a,b)$ are both open and closed – user219081 May 15 '15 at 21:02
• Then we are both right :) This confirms my suspicion that this is the lower limit topology. In the lower limit topology $(-\infty,a)$ is open as is $[b,\infty)$, hence the complement of the union (which is $[a,b)$) must be closed. This is a very interesting space! – graydad May 15 '15 at 21:04

We have now deduced that $[a,b)$ is both open and closed in this topology. Further, sets of the form $(c,d)$ are also open. To see this, let $\left\{[c_i,d)\right\}_{i=1}^\infty$ be a collection of open sets such that $c_i \to c, \space$ $d>c_i>c$ and $c_{i+1}<c_i$. It would be good practice to prove that $$\bigcup_{i=1}^\infty[c_i,d) = (c,d)$$ An arbitrary union of open sets is open, so $(c,d)$ is open. Knowing that $(c,d)$ is open means this topology is a refinement of the standard topology on $\mathbb{R}$. That is to say, open sets in the standard topology are open in this topology (the reverse is not true). From knowing that $(c,d)$ is open, we immediately get that $[c,d]$ is closed in this topology. We may now proceed with the question!

$(1)$ We know $[0,1)$ is open and $(3,4)\subset (3,4]$ is open (largest open set contained in $(3,4]$, so $\text{Int}(A) = [0,1)\cup(3,4)$

$(2)$ We have that $[0,1)$ is closed and $[3,4]$ is the smallest closed set containing $(3,4]$. A union of finitely many closed sets is closed, so $[0,1) \cup [3,4]$ is closed. It is also the smallest closed set containing $[0,1)\cup(3,4]$ so it is the closure.

$(3)$ Use $\text{Ext}(A) = \mathbb{R} \setminus \text{Cl}(A)$

$(4)$ Use $\text{Bd}(A) = \text{Cl}(A) \setminus \text{Int}(A)$

You may want to add a bit more rigor to $(1)$ and $(2)$ besides my claim that such and such a set is the "smallest closed set" or "largest open set" to arrive at an answer.

• so for $Ext(A) = (-\infty, 0) \cup [1,3) \cup (4, \infty)$ and $Bd(A) = \{0,3,4\}$ – user219081 May 15 '15 at 21:32
• @AlyssaWallace yes for exterior. For the boundary, I made a typo on the interior which I fixed. It will change your answer. Also, double check about $3$ being in the boundary – graydad May 15 '15 at 21:37

The above answers for $Cl(A)$, $Int(A)$, and $Ext(A)$ are correct based on my calculations.

$X = \Bbb R \\ Int(A) = [0,1) \cup (3,4) \\ Ext(A) = (-\infty,0) \cup [1,3) \cup (4,\infty) \\ Bd(A) = X \setminus Int(A) \setminus Ext(A) \implies Bd(A) = \{3,4\} .$