Let $X$ be a set and $\left(X, \mathcal{T}_{\mathrm{f}}\right)$ be the topological space with the finite complement topology $\mathcal{T}_{\mathrm{f}}$. Show: For any subset $Y \subseteq X$, the subspace topology on $Y$ inherited from $\left(X, \mathcal{T}_{\mathrm{f}}\right)$ is the same as the finite complement topology on $Y$.
Utilizing the given definitions of the subspace topology and the finite complement topology, we know that a topological space $\left(X, \mathcal{T}_f\right)$ with the finite complement topology $\mathcal{T}_f$, where a subset $U$ of $X$ is open if $X-U$ is finite or if $U=\varnothing$. The subspace topology $\mathcal{T}_Y$ on $Y$, inherited from $\left(X, \mathcal{T}_f\right)$, is defined as $\mathcal{T}_Y=\left\{Y \cap U \mid U \in \mathcal{T}_f\right\}$.
Now, by the definition of subspace and finite complement topologies:
- $\mathcal{T}_f$, the finite complement topology on $X$, is defined as: $$ \mathcal{T}_f=\{U \subseteq X \mid U=\varnothing \text { or } X-U \text { is a finite set }\} $$
- $\mathcal{T}_Y$, the subspace topology on $Y$, is defined as: $$ \mathcal{T}_Y=\left\{Y \cap U \mid U \in \mathcal{T}_f\right\} $$
- $\mathcal{T}_f^Y$, the finite complement topology directly defined on $Y$, is defined as: $$ \mathcal{T}_f^Y=\{V \subseteq Y \mid V=\varnothing \text { or } Y-V \text { is a finite set }\} $$
To prove: $\mathcal{T}_f^Y=\mathcal{T}_Y$ by first proving $\mathcal{T}_Y \subseteq \mathcal{T}_f^Y$ and then $\mathcal{T}_f^Y \subseteq \mathcal{T}_Y$.
1. $\mathcal{T}_Y \subseteq \mathcal{T}_f^Y$
- Consider an arbitrary set $W \in \mathcal{T}_Y$. Then there exists a $U \in \mathcal{T}_f$ such that $W=Y \cap U$.
- Since $U \in \mathcal{T}_f$, we have two possibilities for $U$ :
1.1. $U=\varnothing$, which implies $W=Y \cap \varnothing=\varnothing$, and $\varnothing$ is obviously open in any topology, including $\mathcal{T}_f^Y$.
1.2. $U \neq \varnothing$, and hence $X-U$ is finite since $U$ is open in $\mathcal{T}_f$. This means that the complement of $U$ in $X$ is finite.
- If $U \neq \varnothing$, consider the complement of $W$ in $Y$, given by $Y-W=Y \backslash(Y \cap U)=$ ?
I don't know how to continue the proof.
2. $\mathcal{T}_f^Y \subseteq \mathcal{T}_Y$ :
- Take an arbitrary set $V \in \mathcal{T}_f^Y$. Then $V=\emptyset$ or $Y-V$ is finite, since $V$ is open in the finite complement topology on $Y$.\
1.1. If $V=\emptyset$, then $V \in \mathcal{T}_Y$ since $\mathcal{T}_Y$ is a topology.
1.2. If $Y-V$ is finite.
I also don't know how to go further here. I would appreciate some help.
You have the right idea.
Below, I use "$-$" to denote the set-theoretic difference. We will also assume that $U \ne \emptyset$. If it is the case that $U = \emptyset$, it is trivially open in any topology.
($\mathcal{T}_Y \subseteq \mathcal{T}_f^Y$) Let $U \in \mathcal{T}_Y$. Then by definition, there exists $V \in \mathcal{T}_f^X$ such that $U = V \cap Y$. To show that $U \in \mathcal{T}_f^Y$, we need to show that $Y - U$ is finite. Consider the fact that
$$Y - U = Y \cap (X - U) = Y \cap (X - V \cap Y) \subseteq X - V \cap Y \subseteq X - V$$
Since $X - V$ is finite by hypothesis, $Y - U$ must be finite, as desired.
($\mathcal{T}_f^Y \subseteq \mathcal{T}_Y$) Let $U \in \mathcal{T}_f^Y$; we would like to show that $U \in \mathcal{T}_Y$. To do so, we can find $V \in \mathcal{T}_f^X$ such that $U = Y \cap V$. Consider $V = (X - Y) \cup U$. It is easy to check that $V$ is open in $X$, since
$$X - V = X - ((X - Y) \cup U) = (X - (X - Y)) \cap (X - U) = Y \cap (X - U) = Y - U$$
which is finite by hypothesis. And it is easy to check that
$$Y \cap V = (Y \cap (X - Y)) \cup (Y \cap U) = \emptyset \cup (Y \cap U) = Y \cap U = U$$