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If $X = \mathbb{R}$ with the standard topology and $A = [0, 1)$, then find $\partial A$.

By definition, $\partial A = \overline{A} \cap \overline{X \setminus A}$, where $\overline{A}$ is the closure of $A$. I also know that $\overline{A \cup B} = \overline{A} \cup \overline{B}$.

Obviously, $\overline{A} = [0,1]$. But I'm having trouble dealing with $\overline{X \setminus A}$.

I know that $X \setminus A = (-\infty, 0) \cup [1, \infty)$.

So $\overline{X \setminus A} = \overline{(-\infty, 0) \cup [1,\infty)} = \overline{(-\infty , 0)} \cup \overline{[1, \infty)}$.

I don't know how to continue from here, but it is "obvious" that $\partial A = \{0,1\}$.

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  • $\begingroup$ Draw a picture (you get an interval), what's on the ends of the interval? Two point, $0$ and $1$. $\endgroup$ – IAmNoOne May 12 '16 at 20:05
  • $\begingroup$ Ah, I feel a little silly now - I wanted to write, for example, that $\overline{(-\infty, 0)} = [-\infty, 0]$, which doesn't make much sense at all, when really it should be $(-\infty, 0]$, which is closed. $\endgroup$ – Irregular User May 12 '16 at 20:10
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$\mathrm{cl}((-\infty,0))=(-\infty,0],\mathrm{cl}((1,\infty))=[1,\infty)$.

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$\overline{X \setminus A}$ is $(-\infty,0]\cup[1,+\infty)$, hence $\overline{X\setminus A}\cap\overline{A}=\{0,1\}=\partial{A}$

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$[1, \infty)$ is already closed, so its closure is $[1, \infty)$. As for $(-\infty, 0)$, its closure is $(-\infty, 0]$. This is because $(-\infty, 0)$ is not closed, so $(-\infty, 0)$ is strictly contained in its closure, and adding $0$ to $(-\infty, 0)$ gives a closed set. So: $$\overline{A} \cap \overline{X\backslash A} = [0, 1] \cap ((-\infty, 0] \cup [1, \infty)) = \{0, 1 \}. $$

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What about alternate characterization that $x \in \mathbb{R}$ is a boundary point of $A$ if and only if each neighborhood of $x$ intersects both $A$ and $\mathbb{R} - A$?

Evidently no point of $\mathbb{R} - [0, 1]$ satisfies that condition, nor does any point of $(0, 1)$. But both $0$ and $1$ do satisfy the condition.

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