# Prove or disprove the following statement involving closure, interior, and boundary points

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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As for definitions, there are tons of theorems and definitions we use. My professor doesn't necessarily care what we use to proof with as long as our logic is correct and our proof is followable. – enlgmatlc Mar 30 '11 at 0:16
Please try to use some mark-up; you can see how people do it by right clicking on a formula and then selecting "show source"; you enclose formulas in dollar signs. – Arturo Magidin Mar 30 '11 at 16:14
Will do! Thanks for the info. – enlgmatlc Mar 30 '11 at 22:21

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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Take 2: Proof First, prove cl S is a subset of int S U bd S. Let x belong to cl S. Then x belongs to S U bd S. If x belongs to bd S then we are done. If x belongs to S and x does not belong to bd S then there exists E>0 such that N(x,E) is a subset of S. Therefore, x belongs to int S. Hence, cl S is a subset of int S U 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. – enlgmatlc Mar 30 '11 at 0:12
Looks good, assuming you're allowed to use the results you used. – Alex Becker Mar 30 '11 at 0:39
Awesome, thanks for pointing out my initial mistake, "By the fact int S is a subset of S, x belongs to int S" is false. It helped tons!! – enlgmatlc Mar 30 '11 at 1:36
I'm sorry to be a nuisance, but could you tell me how to use the belong to, such that, square root, ect... symbols? – enlgmatlc Mar 30 '11 at 1:45
Sure. These symbols can be coded using LaTeX, which is built into the math.stackexchange.com site. The code "\in" gives $\in$, "subseteq" gives $\subseteq$, etc. You can find a guide to LaTeX coding by googling "latex commands" or "latex tutorial". – Alex Becker Mar 30 '11 at 1:55

$x$ 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.

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