let Y be a subspace of normed space X . then Y is nowhere dense in X (i.e the interior of closure of Y is empty) iff Y is not dense in X then Y is nowhere dense in X iff Y is closed in X .  

The problem statement presents some logical equivalences about the relationship between a subspace \( Y \) of a normed space \( X \) and its density properties. Let's break this down into a clearer form and address each statement carefully.

### Problem Analysis:
1. **Nowhere Density**: A subset \( Y \) of a topological space \( X \) is called **nowhere dense** if the interior of its closure is empty, i.e., \(\text{int}(\overline{Y}) = \emptyset\). 

2. **Density**: \( Y \) is said to be **dense** in \( X \) if its closure is the entire space, i.e., \(\overline{Y} = X\).

### Statements Given:
1. \( Y \) is nowhere dense in \( X \) **if and only if** \( Y \) is not dense in \( X \).

2. \( Y \) is nowhere dense in \( X \) **if and only if** \( Y \) is closed in \( X \).

### Verification of Statements:

#### 1. **First Statement**:
   - **"Nowhere Dense \(\iff\) Not Dense"**:
     - **(\(\Rightarrow\))**: If \( Y \) is nowhere dense, then \( \text{int}(\overline{Y}) = \emptyset \). Since \( Y \) is nowhere dense, \( \overline{Y} \) cannot be the whole space \( X \), so \( Y \) is not dense in \( X \).
     - **(\(\Leftarrow\))**: If \( Y \) is not dense in \( X \), then \( \overline{Y} \neq X \). This implies there exists some open set in \( X \) that is disjoint from \( \overline{Y} \), which means \( \text{int}(\overline{Y}) \) must be empty, hence \( Y \) is nowhere dense.
     - **Conclusion**: This equivalence is correct.

#### 2. **Second Statement**:
   - **"Nowhere Dense \(\iff\) Closed"**:
     - This statement is generally **not correct**. Here’s why:
     - A set being nowhere dense implies that its closure has empty interior, but this does not necessarily mean the set is closed. A nowhere dense set can be open, closed, or neither.
     - For example, consider the rationals \( \mathbb{Q} \) as a subset of \( \mathbb{R} \) with the standard topology. \( \mathbb{Q} \) is nowhere dense in \( \mathbb{R} \) because \( \overline{\mathbb{Q}} = \mathbb{R} \), and the interior of \( \mathbb{R} \) relative to \( \mathbb{Q} \) is empty. However, \( \mathbb{Q} \) is not closed in \( \mathbb{R} \) (its closure is \( \mathbb{R} \)).
     - Thus, being nowhere dense does not imply the set is closed. Therefore, the second statement is **false**.

### Summary:
- The correct relationship between these properties is:
  - \( Y \) is nowhere dense in \( X \) if and only if \( Y \) is not dense in \( X \).
  - The condition that \( Y \) is nowhere dense does **not** imply that \( Y \) is closed.

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Do you have any questions, or would you like a deeper explanation of any part of this? Here are five related questions that you might find interesting:

1. What is the definition of a dense subset in a normed space?
2. How do we prove that the closure of a dense set in \( X \) is \( X \) itself?
3. Can a set be both open and nowhere dense in a normed space?
4. How does the concept of nowhere density relate to the Baire Category Theorem?
5. What are the implications of a subset being both closed and nowhere dense?

**Tip**: Remember that nowhere dense sets are "small" in a topological sense, which can be useful when considering properties like the Baire Category Theorem.

Explore the logical equivalences between nowhere dense sets and density in normed spaces. Learn about the definitions, implications, and proofs related to these concepts in topology and functional analysis.

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