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Let $C\subset [0,1]$ be the usual Cantor set. Is $[0,a]\cap C$ both open and closed in the relative topology of $C$, whenever $a>0$? Of course it is closed, so the question is about being open.

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If $a\in C$ is not a left endpoint of one of the open intervals that were removed in the construction of $C$, then $[0,a]\cap C$ is not open in $C$. One way to see this is to let $$a=\sum_{i\ge 1}\frac{a_i}{3^i}\,,$$ where each $a_i$ is $0$ or $2$. Since $a$ is not a left endpoint of the removed intervals, for each positive integer $n$ there is $i>n$ such that $a_i=0$. Thus, there is a strictly increasing sequence $\langle n_k:k\in\mathbb{N}\rangle$ of positive integers such that $a_{n_k}=0$ for all $k\in\mathbb{N}$. For $k\in\mathbb{N}$ let $$b_k=\sum_{k=1}^{n_k-1}\frac{a_i}{3^i}+\frac2{3^{n_k}}\,;$$ clearly $b_k\in C$, and

$$\begin{align*}a&=\sum_{i=1}^{n_k-1}\frac{a_i}{3^i}+\frac0{3^{n_k}}+\sum_{i>n_k}\frac{a_i}{3^i}\\ &<\sum_{i=1}^{n_k-1}\frac{a_i}{3^i}+\frac1{3^{n_k}}\\\\\\ &<b_k\\\\ &<\sum_{i=1}^{n_k-1}\frac{a_i}{3^i}+\frac3{3^{n_k}}\\ &=\sum_{i=1}^{n_k-1}\frac{a_i}{3^i}+\frac1{3^{n_k-1}}\;, \end{align*}$$

i.e., $b_k\in\left(a,a+\dfrac1{3^{n_k-1}}\right)$. It follows that $\langle b_k:k\in\mathbb{N}\rangle$ is a sequence in $C\setminus[0,a]$ converging to $a$ and hence that $[0,a]\cap C$ is not open in $C$.

If $a$ is one of the countably many left endpoints of removed intervals, or if $a\in[0,1]\setminus C$, then $[0,a]\cap C$ is open in $C$. This is clear if $a\in[0,1]\setminus C$, as in that case $[0,a]\cap C=[0,a)\cap C$. The case in which $a$ is a left endpoint of some removed open interval $J$ is almost as easy: just take $b$ to be any point of $J$, and $[0,a]\cap C=[0,b)\cap C$.

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