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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.

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    Will do! Thanks for the info.2011-03-30

2 Answers 2

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What assumptions are valid depends on how $\text{bd }S$, $\text{int }S$ and $\text{cl }S$ have been defined in your class, so it would be nice if you could explain this.

Some criticism:

First part: The statement "By the fact $\text{int }S$ is a subset of $S$, $x$ belongs to $\text{int }S$" is false.

Second part: Correct, but much more verbose than necessary, sentences 3 and 4 are unnecessary.

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    Thanks!! You've been great!2011-03-30
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$x$ is in cl $S$ iff every neighbourhood of $x$ intersects $S$. There are 2 mutually exclusive cases: there exists a neighbourhood $U$ that does not intersect $X \setminus S$, or all neighbourhoods of $x$ intersect $X \setminus S$. In the latter case we have exactly that $x$ is in bd $S$, and in the former case $U$ witnesses that $x$ is in int $S$. Both these sets are of course subsets of cl $S$, which finishes the proof.