Interior and Closure in the Standard Topology on [-1,0)U(0,1]

The standard topology of $ Y = [-1,0) \cup (0,1] $ inherits its topology from the real line $ \mathbb{R} $.

In other words, it is a topological subspace of $ \mathbb{R} $ because $ Y \subseteq \mathbb{R} $.

Open Sets in the Topology Y

In a topological subspace, a set $ A \subseteq Y $ is open in $ Y $ if and only if $ A $ can be expressed as the intersection of an open set in $ \mathbb{R} $ with $ Y $.

This means that $ A $ is open in $ Y $ if there exists an open set $ O \subseteq \mathbb{R} $ such that $ A = O \cap Y $.

To better understand which sets are open and closed, and the closure and interior of sets in $ Y $, we need to consider the intersection of open and closed sets of the real line with $ Y $.

For example, $ Y $ itself, i.e., $ [-1,0) \cup (0,1] $, is open because both [-1,0) and (0,1] can be obtained as the intersection of Y with open sets in $ \mathbb{R} $ like (-1.5,0.5) and (-0.5,1.5).

$$ (-1.5,0.5) \cap Y = (-1.5,0.5) \cap ( [-1,0) \cup (0,1] ) = [-1,0) $$

$$ (-0.5,1.5) \cap Y = (-0.5,1.5) \cap ( [-1,0) \cup (0,1] ) = (0,1] $$

Therefore, the sets [-1,0) and (0,1] are both open.

Knowing that the union of open sets is still an open set, the set $ Y = [-1,0) \cup (0,1] $ is also open.

In general, the open sets include:

$ [-1, a) $ for $ -1 < a < 0 $
$ (b, 1] $ for $ 0 < b < 1 $.
$ (-1, a) \cup (b, 1] $ for $ -1 < a < 0 \lt; b < 1 $.

Closed Sets in the Topology Y

Closed sets in $ Y $ are those that can be written as the intersection of a closed set in $ \mathbb{R} $ with $ Y $.

A set $ B \subseteq Y $ is closed in $ Y $ if and only if $ Y \setminus B $ is open in $ Y $.

This means that $ B $ is closed in $ Y $ if it can be written as the intersection of a closed set in $ \mathbb{R} $ with $ Y $.

The set [-1,0) in Y is also closed because it can be obtained as the intersection of the closed set [-1.5,0] in $ \mathbb{R} $ with $ Y $.

$$ [-1.5,0] \cap Y = [-1.5,0] \cap ( [-1,0) \cup (0,1] ) = [-1,0) $$

Note. Alternatively, we can follow this reasoning: A set is closed if its complement is open. In this case, the complement of [-1,0) in the topology Y is the set (0,1], which is open in Y. Therefore, the set [-1,0) is the complement of an open set, meaning [-1,0) is a closed set in Y.

For the same reason, the set (0,1] is also closed in Y because it can be obtained as the intersection between the closed set [0,1] in $ \mathbb{R} $ and Y.

$$ [0,1] \cap Y = [0,1] \cap ( [-1,0) \cup (0,1] ) = (0,1] $$

Therefore, knowing that the union of closed sets is still a closed set, $ Y = [-1,0) \cup (0,1] $ is also a closed set.

Note. Alternatively, we can deduce that Y is closed in its own topology because its complement is the empty set $\emptyset$, which is open in any topology. The complement of an open set is a closed set. Therefore, Y is closed.

In general, closed sets in $ Y $ include:

Y itself.
$ [-1, a) $ for $ -1 < a < 0 $
$ (b, 1] $ for $ 0 < b < 1 $.
$[-1, a) \cup (0, b] $ for $ -1 \le a < 0 $ and $ 0 < b \le 1 $.

Note. The set $\{0\}$ is not closed in $ Y $ because it cannot be written as the intersection of a closed set in $ \mathbb{R} $ with $ Y $.

Closure

The closure of a set $ A \subseteq Y $ is the intersection of the closure of $ A $ in $ \mathbb{R} $ with $ Y $.

$$ \text{Cl}(A) \cap Y $$

If $ \text{Cl}(A) $ denotes the closure of $ A $ in $ \mathbb{R} $, then the closure of $ A $ in $ Y $ is $ \text{Cl}(A) \cap Y $.

For example, the closure of the set [-1,0) in $ \mathbb{R} $ is the set [-1,0].

Therefore, the closure of [-1,0) in Y is the intersection between its closure in $ \mathbb{R} $, which is [-1,0], and the set $ Y = [-1,0) \cup (0,1] $ itself.

$$ [-1,0] \cap Y = [-1,0) $$

For the same reason, the closure of (0,1] in $ \mathbb{R} $ is [0,1], and the closure of (0,1] in Y is still (0,1].

$$ [0,1] \cap Y = (0,1] $$

Therefore, the closure of $ Y = [-1,0) \cup (0,1] $ in $ Y $ is $ [-1,0) \cup (0,1] $.

Interior

The interior of a set $ A \subseteq Y $ is the intersection of the interior of $ A $ in $ \mathbb{R} $ with $ Y $.

If $ \text{Int}(A) $ denotes the interior of $ A $ in $ \mathbb{R} $, then the interior of $ A $ in $ Y $ is $ \text{Int}(A) \cap Y $.

For example, the interior of [-1,0) in $ \mathbb{R} $ is (-1,0). Therefore, the interior of [-1,0) in Y is the intersection between (-1,0) and Y.

$$ (-1,0) \cap Y = (-1,0) $$

The interior of (0,1] in $ \mathbb{R} $ is (0,1). Therefore, the interior of (0,1] in Y is the intersection between (0,1) and Y.

$$ (0,1) \cap Y = (0,1) $$

The interior of $ Y = [-1,0) \cup (0,1] $ in $ Y $ is $ (-1,0) \cup (0,1) $, so the interior of Y is the intersection between $ (-1,0) \cup (0,1) $ and Y.

$$ [ (-1,0) \cup (0,1) ] \cap Y = (-1,0) \cup (0,1) $$

And so on.