# Defining a topology on the middle third Cantor set $C$

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
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

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.