Exercise on the Interior and Closure of a Set in Topology 5

In this exercise, we need to find the interior and closure of the set \(A = (0,1) \cup \{2\}\) within the standard topology on \(\mathbb{R}\):

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

For the set \(A\), the only open interval that can be fully contained in \(A\) is \((0,1)\).

The point \(2\) is isolated and does not belong to any open interval within \(A\).

Thus, the interior of \(A\) is:

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

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

Limit points of \(A\) are those points that can be approached by points within \(A\).

The interval \((0,1)\) is already open and includes all its limit points between 0 and 1.

The closure of \(A\) also includes the boundary points 0 and 1, as any interval containing 0 or 1 contains points from \(A\).

The point \(2\) is isolated and has no limit points but must be included in the closure.

Therefore, the closure of \(A\) is:

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

In summary:

The interior of \(A = (0,1) \cup \{2\}\) is \(\text{Int}(A) = (0,1)\).

The closure of \(A = (0,1) \cup \{2\}\) is \(\text{Cl}(A) = [0,1] \cup \{2\}\).

And so on.