Exercise on the Interior and Closure of a Set in Topology 6

In this exercise, we need to determine the interior and closure of the set $A = (0,1) \cup \{2\}$ in $\mathbb{R}$ under the lower limit topology.

The interior of a set is the union of all open subsets contained within it.

In the lower limit topology, open sets are of the form $[a, b)$ for any $a < b$.

The interval $(0,1)$ is not open in this topology because it cannot be expressed as a union of intervals of the type $[a, b)$.

However, all intervals of the form [a,1) that are contained within (0,1) and where a>0 are open, and the union of these intervals is (0,1).

$$ \text{Int}(A) = \bigcup_{0 < a < b < 1} [a, b) $$

The point $ \{2\} $ is isolated and thus does not contribute to the interior.

Therefore, the interior of $A$ is the interval (0,1).

\[ \text{Int}(A) = (0,1) \]

The closure of a set is the union of the set itself and its set of accumulation points.

In the lower limit topology, the accumulation points of a set $A$ are those points where every interval of the form $[a, b)$ intersects $A$.

The interval $(0,1)$ has a closure of $[0,1]$ in this topology because every open interval $[a, b)$ with $a \leq 0 < 1 \leq b$ intersects $(0,1)$.

The point \{2\} is isolated, so it must be included in the closure but does not introduce any new accumulation points.

Therefore, the closure of $A$ is:

\[ \text{Cl}(A) = [0,1] \cup \{2\} \]

The interior of $A = (0,1) \cup \{2\}$ in the lower limit topology is $\text{Int}(A) = (0,1) $.

The closure of $A = (0,1) \cup \{2\}$ in the lower limit topology is $\text{Cl}(A) = [0,1] \cup \{2\}$.

And so forth.