# the cofinite topology in $Y$ and the topology of $Y$-subspace induced by the topology of $X$

Be $$X$$ a set of cofinite topology and $$Y \subseteq X$$. Are $$\tau_{c}$$ and $$\tau_{s}$$, respectively, the cofinite topology in $$Y$$ and the topology of $$Y$$-subspace induced by the topology of $$X$$. Compare $$\tau_{c}$$ and $$\tau_{s}$$ with respect inclusion.

Definition 1. Be $$X$$ an infinite set and set $$\tau=\{\emptyset\}\cup \{V\subseteq X:X-V$$ is finite$$\}$$.

Definition 2. Be $$X$$ a set, $$\tau$$ a topology, and $$Y\subseteq X$$. Then the $$X$$-induced topology is given by: take the $$X$$ open and intersect with $$Y$$.

Can anyone help me with this? I think the only difference from $$\tau_{c}$$ topology and topology is that in $$\tau_{c}$$ topology: $$\begin{equation} \tau_{c}=\{\emptyset\}\cup \{V\subseteq Y: Y-V \text{ is finite } \} \end{equation}$$

while whereas

$$\begin{equation} \tau_{s}=\{\emptyset\}\cup \{V\subseteq X: X-V \text{ is finite } \}\cap Y \end{equation}$$

in my view it doesn't change anything. Change anything, which is the finest?

• You are correct: they are the same topology. – Brian M. Scott Dec 22 '20 at 19:33
• Brian, How can I prove that fact? – Isadora Suhadolnick Dec 22 '20 at 19:39

Take a finite set $$V\subset Y$$, this means $$Y- V$$ is open in the cofinite topology of $$Y$$, now $$X-V$$ is also open in the cofinite topology of $$X$$ and $$Y\cap (X-V)=Y-V$$, so $$Y- V$$ is also open in the topology inherited by $$X$$. It is pretty obvious that $$Y\cap (X- V)$$ is open in the cofinite topology of $$Y$$ for any finite $$V\subset X$$, so you have the double incusion.
If $$V\in\tau_c$$ let $$F=Y\setminus V$$, and let $$U=X\setminus F$$. $$F$$ is a finite subset of $$X$$, so $$U$$ is open in $$X$$, and clearly
$$U\cap Y=(X\setminus F)\cap Y=(X\cap Y)\setminus F=Y\setminus F=V\,,$$
so $$V\in\tau_s$$. This shows that $$\tau_c\subseteq\tau_s$$; now see if you can show that $$\tau_s\subseteq\tau_c$$ to conclude that $$\tau_s=\tau_c$$.