A variation on the induced topology (subspace topology) One of the easiest ways to find a topology on a set is to used the Induced Topology.
Induced Topology: Suppose you have a set $X$ and a topology on $X$ called $\mathcal{T}_X$. Suppose $Y \subseteq X.$ Then we can find a topology for $Y$ (call it $\mathcal{T}_Y$) as given by:
$\mathcal{T}_Y=\{O \cap Y  \mid O \in \mathcal{T}_X\}$.
My question is: does the induced topology still hold for sets $Y \not \subseteq X.$ Suppose $Y \cap X = \emptyset.$ Can we still inherit a topology on $Y$ if $X$ and $Y$ are disjoint? I'm going to say no.
Here's my test:
Let $X =\{1,2\}$ and $\mathcal{T}_X = \{\emptyset,\{1\},\{1,2\}\}$
Let $Y = \{5\}$
Then, following from our induced topology method,
$\mathcal{T}_Y=\{\{5\} \cap \emptyset, \{5\} \cap \{1\}, \{5\} \cap \{1,2\}\} = \{\emptyset,\emptyset,\emptyset\}$
But this fails to satisfy definition of topology since $Y=\{5\}$ and $Y \not \in \mathcal{T}_Y.$
So it would seem that we need $Y \subseteq X$ so that $Y \in \mathcal{T}_Y$, otherwise we end up with a bunch of empty sets.
Is my reasoning here correct?
 A: No, this concept makes no sense unless $Y \subseteq X$. It will only be a topology if this is the case: We must have $Y \in \mathcal{T}_Y$ so $\exists O \in \mathcal{T}_X: O \cap Y = Y$ from which it follows that $Y \subseteq O \subseteq X$.
Note that the definition of $\mathcal{T}_Y$ is not an arbitrary one, it's the natural choice in that it is the smallest topology on $Y$ that makes the inclusion map $i: Y \to X$ continuous. Extending it to any $Y$, unrelated to $X$ does not solve such a natural problem, or is not especially useful in constructing new spaces. So don't consider it, it won't get accepted.
A: You can define
$\tau_Y=\{O\cap Y: O\in\tau_X\}\cup\{Y\}.$
This is a topology on $Y$. It is the topology generated by $\{O\cap Y: O\in\tau_X\}$, i.e. the smallest topolgy containing that family of sets.
(And in fact that topology carries informations only about $Y\cap X$.)
Basically you have a topology on the subset $Y\cap X$, and you promote it to a topology on $Y$ by adding $\{Y\}$.
If $Y\cap X=\emptyset$, then $\tau_Y=\{\emptyset, Y\}$ is just the trivial topology.
