# The relative topology with respect to two spaces is the same.

In Bert Mendelson's "Introduction to Topology" p. 159, i read the statement "A topological space $C$ can be a subspace of two distinct topological spaces $X$ and $Y$. In this event the relative topology of $C$ is the same whether we regard $C$ as a subspace of $X$ or $Y$."

If $J$ is the topology of $X$ and $I$ the topology of $Y$, then the relative topology of $C$ with respect to $X$ consists of sets of the form $C \cap O$, where $O \in J$. Similarly the relative topology with respect to $Y$ consists of sets $C \cap O'$ with $O' \in I$. I don't see why these two relative topologies are identical.

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They must be identical: each of them is the given topology on $C$. If they were different, we couldn't talk about $C$ as '[a] topological space'. –  Brian M. Scott May 6 '12 at 22:07
I still don't understand. Can we prove that they are identical? –  Manos May 6 '12 at 22:18
Hang on, and I'll write up a longer explanation as an answer. –  Brian M. Scott May 6 '12 at 22:21

When we say that a topological space $\langle C,\tau_0\rangle$ is a subspace of a space $\langle X,\tau_1\rangle$, we mean that $C\subseteq X$ and $\tau_0=\{U\cap C:U\in\tau_1\}$. If $\langle C,\tau_0\rangle$ is also a subspace of $\langle Y,\tau_2\rangle$, then by definition it's also true that $C\subseteq Y$ and $\tau_0=\{U\cap C:U\in\tau_2\}$. Thus, if $C$ is a subspace of both $X$ and $Y$, we necessarily have $\{U\cap C:U\in\tau_1\}=\tau_0=\{U\cap C:U\in\tau_2\}$. It's really just a matter of definition, specifically, the definition of subspace.
If $\{U\cap C:U\in\tau_1\}$ and $\{U\cap C:U\in\tau_2\}$ were unequal, we couldn't talk about a single space $C$ that was a subspace of both $X$ and $Y$.
Thanks. Let me see if i understand now: if $C$ is just a subset of both $X$ and $Y$, then it does not mean that the two relative topologies are equal, right? –  Manos May 6 '12 at 22:30
Right. For example, the set of rationals is a subset of the reals. If I give the reals the usual topology, $\Bbb Q$ also gets its usual topology as a subspace. But if I give $\Bbb R$ the discrete topology, $\Bbb Q$ gets the discrete topology as its subspace topology. But $\Bbb Q$ with a particular topology can't be a subspace of both of those at the same time. –  Brian M. Scott May 6 '12 at 22:32