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Is there some sort of a topology defined on the ternary Cantor set $C$ if we consider that $(a,b) \cap C$ is considered to be an open set? Is this the right approach to be taken if wanting to establish a topology on $C$? Hopefully, I will learn more about this in the Topology course I will be taking at my university this semester.

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If $(X,\mathcal T)$ is a topological space and $A\subset X$, you can put a topology $\mathcal T_A$ on $X$, considering the sets $A\cap O$ where $O$ is an open subset of $(X,\mathcal T)$. –  Davide Giraudo Aug 23 '11 at 18:02
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You may find it useful to know that the topology you suggest on the Cantor set (the subspace topology) is homeomorphic to the product topology on the space $\{0,1\}^{\mathbb N}$ where $\{0,1\}$ is discrete. –  Andres Caicedo Aug 23 '11 at 21:30

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That is exactly the approach taken to define the subspace topology on $C$ as a subspace of the real line with its usual topology. Start with the fact that the open intervals form a base for the usual topology on $\mathbb{R}$: a subset of $\mathbb{R}$ is open if and only if it’s a union of sets of the form $(a,b)$. Now transfer this ‘down’ to $C$: declare the sets of the form $(a,b) \cap C$ to be the basic open sets in $C$, and say that a subset of $C$ is open in the subspace topology on $C$ if and only if it’s a union of basic open sets, i.e., of sets of the form $(a,b) \cap C$.

However, you can just as well get all of the open sets in the subspace topology on $C$ at once, without using open intervals. We’ve just said that if $V \subseteq C$ is open in the subspace topology on $C$, there is some family $\mathscr{I}$ of open intervals in $\mathbb{R}$ such that $$V = \bigcup\{I \cap C:I \in \mathscr{I}\} = C \cap \bigcup\{I:I \in \mathscr{I}\} = C \cap \bigcup\mathscr{I}.$$ $\bigcup\mathscr{I}$ is a union of open intervals in $\mathbb{R}$, so it’s just some open set $W$ in $\mathbb{R}$. Thus, every subset $V$ of $C$ that is open in the subspace topology on $C$ is of the form $W \cap C$ for some open set $W$ in $\mathbb{R}$. Conversely, if you start with some open set $W$ in $\mathbb{R}$, you can write it as $W = \bigcup\mathscr{I}$ for some family $\mathscr{I}$ of open intervals, and carry out the same calculation in reverse to see that $W \cap C$ is a union of sets of the form $(a,b) \cap C$.

The sets of the form $(a,b) \cap C$ are the nicest open sets in the subspace topology on $C$, and the easiest to visualize, and they’re enough to give you all of the open sets in the subspace topology on $C$ by taking unions, but you can get all of the open sets at once by taking the sets of the form $W \cap C$ where $W$ is any open set in $\mathbb{R}$, not just an open interval.

As Davide Giraudo said in his comment, one can do essentially the same thing with any subset of any topological space, and you should indeed learn more about this in your topology course.

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