Opsure and Clinterior of subsets of $\mathbb{R}$

Question

We are considering the made up words opsure and clinterior, where the opsure of a set $A$ is the smallest open set containing $A$ and the clinterior of a set $B$ is the largest closed set contained in $B$.

I am tasked with the following:

Give an example a set $X$ which does not have an opsure, and give an example of a set $Y$ which does not have a clinterior. What is the key reason that every set has a closure but not an opsure? What is the key reason that every set has an interior but not a clinterior?"

First we define an open set:

Given a set $X \subset \mathbb R$, $X$ is called open if $\forall_x \in X$, $x$ has an $\epsilon$-neighborhood which is contained in $X$. So an example of an open set is $(a,b)$. This is true $\forall_{a,b} \in \mathbb R$ given $a \neq b$. Additionally, $\emptyset$ and $\mathbb R$ are open sets.

Likewise, we define a closed set:

A set $Y$ is closed if it contains all its limit points. Or, we can say $Y$ is closed if and only if $Y^c$ is open. Examples of closed sets are $[c,d]$ and $\emptyset$ (it is only one of two sets that are both empty and closed! The other one being $\mathbb R$). Again, we assume $\forall_{c,d} \in \mathbb R$ and $c \neq d$.

For the example of a set lacking the property of opsure:

Given an open set $A$, the smallest open set containing $A$ is $A$ itself. So a set lacking opsure, call it $X$, is a set that is not the smallest set that contains itself. An example of this set would be $X = \mathbb R + (0,1)$ since $\mathbb R$ already contains the set $(0,1)$, thus the set $X$ lacks having the property of opsure as $X$ is not the smallest set that contains itself.

Here I am struggling though. I am not sure if my above example works/is correct and I am having a hard time finding a set $Y$ that satisfies the property of having what is defined as a clinterior.

Any help would be greatly appreciated, thanks!

Answer 1

Hint

For a set that has no oposure, consider $[0,1].$ For a set that has no clinterior, consider $(0,1)$.

The "key reason" has to do with the fact that intersections of open sets are not necessarily open and unions of closed sets are not necessarily closed.

Answer 2

I assume that you define smaller as $A$ is smaller than $B$ if $A\subsetneq B$ and the smallest set of a family is a set that is smaller than any other set in the family.

You should probably start with the reason a set always have an interior and closure to see where this reason breaks down for clinterior and opsure.

The reason we have a interior and closure lies in how we can construct them. The interior of a set is the union of it's open subsets and the closure is the intersection of it's closed supersets. The existence is guaranteed since the union of open sets is open and intersection of closed sets is closed.

Now if we would like to do the similar construct we would try intersection of open sets or union of closed sets which doesn't need to be open or closed respectively.