# Proof that subspace topology is independent of 'parent' space

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?

• By definition, a topological subspace is endowed with the subspace topology. If $C$ is a subspace of $X$ and a subspace of $Y$, then the subspace topologies agree by definition – Ben W Dec 21 '18 at 15:32
• What is the complete formulation? E.g. One exact formulation could be: Let $X$ be a topological space, let $Y$ be a subset of $X$ in the subspace topology (from $X$). Let $C$ be a subset of $Y$. Then the subspace topology that $C$ gets as a subspace of $Y$ or of $X$ is the same. This follows easily from the definitions. – Henno Brandsma Dec 21 '18 at 15:33
• I think he means X and Y are common subspaces of a larger space...But then X∩Y is a subspace of X containing C. Using @HennoBrandsma reasoning, we are done. – YuiTo Cheng Dec 21 '18 at 15:33
• @HennoBrandsma I have indeed seen a proof of that case, but paraphrasing the book "A topological space $C$ may be a subspace of two distinct larger topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard..." – Steven Wagter Dec 21 '18 at 15:36
• So the one is not necessarily a subspace of the other. – Steven Wagter Dec 21 '18 at 15:37

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.
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 $$\{x\cap C:x\in T_X\}=T_C=\{y\cap C: y\in T_Y\}.$$