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Problem. Let $\mathcal{M}$ be a topology on a set $X$ and $Y \subset X$. Then prove that $\mathcal{M}_Y = \{Y \cap O | O \in \mathcal{M}$} is defined to be the induced topology on $Y$.

I have to satisfy three axioms (open axioms). One of which is the union of open subsets. Formally, if $Q_i \in \mathcal{M}$ ($\mathcal{M}$ is a topology on $Y$) is open for all $i$ in some interval $I$ then $\bigcup_{i \in I} Q_i \in \mathcal{M}$.

I have proved this, but I think mine is a bit too hand-wavy. Any tips/help to make it more clear and or any glaring errors I didn't notice?

Proof: Let $Z \subset \mathcal{M}_Y$ and denote $\bigcup Z$ as the collection of all open sets such that $\bigcup Z \in \mathcal{M}_Y$. Set \begin{equation*} Z' = \left\{M \in \mathcal{M}~|~M \cap Y \subset \bigcup Z\right\}. \end{equation*} By definition of topology, $Z' \subset \mathcal{M}$. Now, let $O = \bigcup Z'$. Hence, by definition, $O \in \mathcal{M}$. Thus, for $M \in Z'$, \begin{equation*} O \cap Y = \left(\bigcup M \cap Y\right) \subset \bigcup Z. \end{equation*}

Now for all $N \in Z$ there exists a set $M \in \mathcal{M}$ such that $N = M \cap Y \subset \bigcup Z$. But then we also have the for all $N \in Z$ there exists a $M \in Z'$ such that $N = M \cap Y$. Then $M \subset O$, and consequently, $ N \subset O \cap Y$. Hence, it follows that $\bigcup Z \subset O \cap Y$. Therefore, $\bigcup Z = O \cap Y \in \mathcal{M}_{Y}$.

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  • $\begingroup$ Thanks, fixed it. $\endgroup$ – user334916 Sep 22 '17 at 8:55
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    $\begingroup$ Your denoting Z as a collection is stated wrongly and is incorrect for assuming Z is countable. You are only showing that M_y is a topology. No where have you used or refered to the definition of an induced topology to show that M_y is the induced topology. $\endgroup$ – William Elliot Sep 22 '17 at 9:15
  • $\begingroup$ You have $Z ⊂ \mathcal{M}_Y$ and you want to show $⋃Z ∈ \mathcal{M}_Y$. I don't understand what your sets $Z_1, …, Z_n$ are. $\endgroup$ – user87690 Sep 22 '17 at 9:16
  • $\begingroup$ I've edited the proof, and removed unnecessary/sloppy notation. $\endgroup$ – user334916 Sep 22 '17 at 10:22
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Suppose $\mathcal{O} \subseteq \mathcal{M}_Y$ be a collection of subsets in the subspace topology (as defined). So for all $O \in \mathcal{O}$, there is some $\hat{O} \in \mathcal{M}$ such that $\hat{O} \cap Y =O$. Then

$$\bigcup \mathcal{O} = \bigcup \{\hat{O} \cap Y : O \in \mathcal{O}\} = \bigcup \{\hat{O} : O \in \mathcal{O}\} \cap Y \in \mathcal{M}_Y$$

using distributivity of $\cap$ over unions, and the fact that the union of the $\hat{O}$ is open in $\mathcal{M}$.

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If $(X, \mathcal{T})$ is a topological space, and $Y \subseteq X$ is a subset of $X$, then the subspace/induced topology $\mathcal{T}_Y$ on $Y$ is defined to be $$\mathcal{T}_Y = \{ Y \cap U \ | U \in \mathcal{T}\}$$

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