Proof that subspace topology is independent of 'parent' space

Question

In my topology textbook (Bert Mendelson's) it is stated that if $C$ is a subspace of two distinct larger spaces $X$ and $Y$, then the relative topology of $C$ is the same whether we regard it as a subspace of $X$ or $Y$.

In an attempt to prove this, regard $C$ as a subspace of $X$ and suppose $S$ is an open subset of $C$. Then $$S = S' \cap C$$ for some open subset $S'$ of $X$. Now I'm not sure how to continue; proving that $S' \cap Y$ is an open subset of $Y$ would do the job, since then $$S' \cap Y \cap C = S$$ is in the topology of $C$ regarded as a subspace of $Y$, but I don't see a way to prove this.

Can anyone help me with this proof? Can $S' \cap Y$ be shown to be open in $Y$ or should I take a completely different approach?

Thanks in advance!

Answer 1

Following the discussion in the comments I'll assume the following situation: We have a superspace $Z$ such that $X,Y \subseteq Z$ have the subspace topology w.r.t. $Z$ and $C \subseteq X \cap Y$.

Now $C$ can inherit the subspace topology of $X$ or of $Y$, but this does not matter, because in the end it's just the subspace topology from $Z$.

If $S \subseteq C$ is open "via $X$" then $S = S_X \cap C$ with $S_X$ open in $X$, so really, $S_X = O \cap X$ where $O$ is open in $Z$. Hence $S = (O \cap X) \cap C = O \cap C$ (as $X \cap C = C$) and so $S$ is open as a subspace of $Z$. Moreover, $S = O \cap C = O \cap (C \cap Y) = (O \cap Y) \cap C$ where $O \cap Y$ is open in $Y$ (by definition of the subspace topology on $Y$) and so $S$ is also open "via $Y$". This argument is entirely symmetrical, so indeed it does not matter via which subspace $X$ or $Y$ we endow $C$ with a subspace topology. The common superspace enforces the consistency, as it were.

Answer 2

We almost always talk of a top. space $X$ when we really mean a pair $(X,T_X)$ where $T_X$ is a certain kind of collection of subsets of $X$, called a topology on $X$, and commonly called the collection of open sets. $T_X$ is not determined by $X$ alone, unless $X$ has at most one member. Suppose $T_X$ is a topology on $X$ and $T_Y$ is a topology on $Y,$ and that $X\supset C\subset Y.$ There may be many possible topologies on the set $C$. Suppose $T_C$ is a topology on $C$ such that $(C,T_C)$ is a sub-space of $(X,T_X)$ and also of $(Y,T_Y)$. By the definition of a sub-space topology, this means that $$\{x\cap C:x\in T_X\}=T_C=\{y\cap C: y\in T_Y\}.$$