Let $S$ be a subset of $R$ (the set of Real numbers). Prove/disprove: $\text{cl }S = \text{int }S \cup \text{bd }S$. The following is my proof, please help me critique it and fix any incorrect assumptions. Proof:
First, prove $\text{cl }S$ is a subset of $\text{int }S \cup \text{bd }S$ by cases. Case 1: Let $x$ belong to $\text{bd }S$, we are done. Case 2: Let $x$ belong to $\text{cl }S$, but $x$ does not belong to $\text{bd }S$. Then $x\in S$. By the fact $\text{int }S$ is a subset of $S$, $x$ belongs to $\text{int }S$. Therefore $\text{cl }S$ is a subset of $\text{int }S \cup \text{bd }S$.
Second, prove $\text{int }S \cup \text{bd }S$ is a subset of $\text{cl }S$. Assume $x$ does not belong to $\text{cl }S$. Then $x$ does not belong to $S \cup \text{bd }S$. Hence, $x$ does not belong to $S$ and $x$ does not belong to $\text{bd }S$. Since $x$ does not belong to $S$, then $x$ does not belong to $\text{int }S$. Hence $x$ does not belong to $\text{bd }S$ and $x$ does not belong to $\text{int }S$. Therefore $x$ does not belong to $\text{int }S \cup \text{bd }S$. Hence $\text{int }S \cup \text{bd }S$ is a subset of $\text{cl }S$. Therefore $\text{cl }S = \text{int }S \cup \text{bd }S$.
2nd Attempt:
Proof
First, prove $\text{cl } S$ is a subset of $\text{int } S \cup \text{bd }S$.
Let $x\in cl S$. Then $x\in \text{S } \cup \text{bd }S$.
If $x\in \text{bd }S$ then we are done.
If $x\in S$ and x does not belong to $\text{bd }S$ then there exists E>0 such that N(x,E) is a subset of S.
Therefore, $x\in \text{int }S$.
Hence, $\text{cl }S$ is a subset of $\text{int }S \cup \text{bd }S$.
As for the other half of my proof, I don't see anything incorrect at this point in time( do tell if you see anything!!) Please let me know if I need to provide anything further.