Hopefully this question is appropriate for this forum. If not please let me know. I would like to know if the justifications (proofs) of the solution are correct.
Let $E = \{1, 1/2, 1/3, 1/4, \ldots\}$. Also $\sup E = 1$ with $1 \in E$ and $\inf E = 0$ so $E \subset (0,1]$.
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Set of interior points of E: let $c > 0, c \in \mathbb{R}$. Take any $x \in E$. Consider the interval $(x-c,x+c)$ since that interval contains rational and irrational number then $(x-c,x+c) \not\subset E$ so x not interior point of E. Therefore no interior points in E and set of interior point of E is the empty set $\phi$.
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Set of accumulation points of E: for every $c > 0, c \in \mathbb{R}$, take any $x \in (0,1]$. Consider the interval $(x-c,x+c)$ since that interval contains rational and irrational number and $E \subset \mathbb{Q}$ then $(x-c,x+c)\cap E$ contains infinitely many points of E. Therefore the set of accumulation points of E is (0,1].
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Set of isolated points of E: let $c > 0, c \in \mathbb{R}$. Take any $x \in E$. Consider the interval $(x-c,x+c)$ since that interval contains rational and irrational number then $(x-c,x+c) \cap E \subset E \neq \{ x \}$ so x not an isolated point of E. Therefore no isolated points in E and set of isolated points of E is the empty set $\phi$.
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Set of boundary points of E: For every $c > 0, c \in \mathbb{R}$ then every interval $(0-c,0+c)$ has at least one point outside E and at least one point inside E. Also every interval $(1-c,1+c)$ has at least one point outside E and at least one point inside E. Otherwise for any $x \in E$ not every interval $(x-c,x+c)$ has at least one point outside E and at least one point inside E. Therefore the set of boundary points of E is {0,1}.
Note: Reference for definitions of interior, accumulation, isolated and boundary points is "Elementary Real Analysis" by B. Thomson, J.B. Brucker and A.M. Bruckner, Sec. 4.2, p. 165.
Thanks in advance for the comments.
Best Answer
It's correct to say that $(x - c, x + c)$ contains both rational and irrational numbers, and it's correct to say that $E \subseteq \mathbb{Q}$; but it's not correct to say that $(x - c, x + c)$ contains some member of $E$ as a consequence. As a simple example, let $x = 3/4$ and $c = 1/8$; the interval $(5/8,7/8)$ contains no members of $E$. Importantly, just because all members of $E$ are in $\mathbb{Q}$ does not mean that the members of $\mathbb{Q}$ in $(x - c, x + c)$ happen to be the same ones that are in $E$!
This error affects your answers in 2, 3, and 4. To get you started on fixing it, here's a suggestion about #2.
Let $1/2 < x < 1$. The interval $(1/2, 1)$ is an open interval containing $x$ which does not include any member of $E$ (since all members of $E$ other than $1$ and $1/2$ are less than $1/2$), so $x$ is not an accumulation point of $E$.
I'll leave it to you, for now, to apply this line of thinking more generally.