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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?

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  • $\begingroup$ Your reasoning is correct, but $\{5 \cap O : O \in \mathcal T_X\}$ should be $\{\{5\} \cap O : O \in \mathcal T_X\}$. $\endgroup$ – azif00 Feb 16 at 1:38
  • $\begingroup$ Ah you're correct. I gotta watch my levels of existence. I'll fix it $\endgroup$ – Emily Burkenhamen Feb 16 at 1:39
  • $\begingroup$ If you make some random extension to.the universe of discourse of a mathematical structure, why would you expect it to have any good properties? $\endgroup$ – Rob Arthan Feb 16 at 1:47
  • $\begingroup$ @RobArthan are you saying that nothing would be special about topologies if we could just construct one from a known one, even if $X$ and $Y$ have nothing to do with each other? $\endgroup$ – Emily Burkenhamen Feb 16 at 2:03
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    $\begingroup$ @EmilyBurkenhamen I find Rob Arthan's comments entirely unhelpful, and I hope you're not discouraged by them. Welcome to MathSE, where that sort of this is disappointingly common. I think your question has merit, and is well phrased. I also think that your response to Rob's comment made significantly more sense than his own first comment. Whilst Rob apparently didn't understand your response (despite the fact that it was less jam-packed with mathematical psychobabble than his own comment), I did. The answer is yes, that is exactly what he's saying. $\endgroup$ – Matt Feb 16 at 11:28
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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.

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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.

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