[proofplan] We prove equality of the two topologies on $Y$ by proving both inclusions. A subspace-[open set](/page/Open%20Set) has the form $Y\cap U$ with $U$ cofinite-open in $X$, and its complement in $Y$ is controlled by the finite set $X\setminus U$. Conversely, a cofinite-open subset of $Y$ has finite complement $F$ in $Y$, and the open set $X\setminus F$ in $X$ restricts exactly to it. [/proofplan] [step:Show every subspace-open set is cofinite-open in $Y$] Let $\tau_Y$ denote the [subspace topology](/page/Subspace%20Topology) on $Y$. Let $V\in \tau_Y$. By the definition of the subspace topology, there exists an open set $U\in \tau_{\mathrm{cof}}(X)$ such that $V=Y\cap U$. If $U=\varnothing$, then $V=\varnothing$, so $V\in\tau_{\mathrm{cof}}(Y)$. Suppose instead that $U\neq\varnothing$. Since $U$ is open in the [cofinite topology](/page/Cofinite%20Topology) on $X$, the complement $X\setminus U$ is finite. The complement of $V$ in $Y$ is \begin{align*} Y\setminus V = Y\setminus (Y\cap U). \end{align*} Using the set identity \begin{align*} Y\setminus (Y\cap U) = Y\cap (X\setminus U). \end{align*} we obtain \begin{align*} Y\setminus V = Y\cap (X\setminus U). \end{align*} Because $X\setminus U$ is finite, its subset $Y\cap (X\setminus U)$ is finite. Hence $Y\setminus V$ is finite, so $V$ is open in the cofinite topology on $Y$. Therefore $\tau_Y\subset \tau_{\mathrm{cof}}(Y)$. [guided] Let $\tau_Y$ denote the topology on $Y$ obtained as the subspace topology from $X$. To prove that every subspace-open set is cofinite-open in $Y$, take an arbitrary set $V\in\tau_Y$. The definition of the subspace topology says that there is some open set $U\in\tau_{\mathrm{cof}}(X)$ with $V=Y\cap U$. We now use the definition of the cofinite topology on $X$. There are two cases. If $U=\varnothing$, then $V=Y\cap \varnothing=\varnothing$, and $\varnothing$ is open in the cofinite topology on $Y$ by definition. Now suppose $U\neq\varnothing$. Since $U$ is open in the cofinite topology on $X$, its complement $X\setminus U$ is finite. We must show that $V$ is cofinite-open in $Y$, which means that $V=\varnothing$ or $Y\setminus V$ is finite. We compute the complement of $V$ inside the ambient set $Y$: $Y\setminus V=Y\setminus (Y\cap U)$. By the elementary set identity $Y\setminus (Y\cap U)=Y\cap (X\setminus U)$, this becomes $Y\setminus V=Y\cap (X\setminus U)$. The set $Y\cap (X\setminus U)$ is a subset of the finite set $X\setminus U$, so it is finite. Hence $Y\setminus V$ is finite, and therefore $V$ is open in the cofinite topology on $Y$. Since the argument applies to every $V\in\tau_Y$, we have proved $\tau_Y\subset \tau_{\mathrm{cof}}(Y)$. [/guided] [/step] [step:Show every cofinite-open set in $Y$ is subspace-open] Let $V\in\tau_{\mathrm{cof}}(Y)$. If $V=\varnothing$, then \begin{align*} V=Y\cap \varnothing, \end{align*} and $\varnothing$ is open in $X$, so $V\in\tau_Y$. Suppose instead that $V\neq\varnothing$. Since $V$ is open in the cofinite topology on $Y$, the set $F:=Y\setminus V$ is finite. Define $U:=X\setminus F$. Because $F$ is finite and $F\subset X$, the set $U$ is open in the cofinite topology on $X$. Moreover, since $F=Y\setminus V$ and $V\subset Y$, we have $Y\cap U=Y\cap (X\setminus F)=Y\setminus F=V$. Thus $V$ is the intersection of $Y$ with an open set of $X$, so $V\in\tau_Y$. Therefore $\tau_{\mathrm{cof}}(Y)\subset \tau_Y$. [/step] [step:Conclude the two topologies are equal] The first step proved \begin{align*} \tau_Y\subset \tau_{\mathrm{cof}}(Y), \end{align*} and the second step proved $\tau_{\mathrm{cof}}(Y)\subset \tau_Y$. Hence $\tau_Y=\tau_{\mathrm{cof}}(Y)$. Therefore the topology on $Y$ induced as a subspace of the cofinite space $X$ is exactly the cofinite topology on $Y$. [/step]