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If $(X,\tau)$ is a topological space and $A \not\in \tau$ , then $\tau(A) = \{U_1 \cup (U_2 \cap A) : U_1, U_2 \in \tau\} $ is the topology on $X$ generated by the subbase $\{A\} \cup \tau$.

Let $(X, \tau)$ be Lindelöf (or sequential compact) and $A \subset X$ be Lindelöf (or sequential compact). Is $(X,\tau(X - A))$ Lindelöf (or sequential compact)? why?

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The answer is yes in both cases.

Suppose that $\langle X,\tau\rangle$ is Lindelöf, and that $A\subseteq X$ is also Lindelöf, and let $\mathscr{U}$ be a $\tau(X\setminus A)$-open cover of $X$. For each $U\in\mathscr{U}$ let $V_U,W_U\in\tau$ be such that $$U=V_U\cup\big(W_U\cap(X\setminus A)\big)=V_U\cup(W_U\setminus A)\;.$$

$A$ is Lindelöf, and $\tau$ and $\tau(X\setminus A)$ generate the same relative topologies on $X\setminus A$, so there is a countable $\mathscr{U}_0\subseteq\mathscr{U}$ such that $A\subseteq\bigcup\mathscr{U}_0$. Let $V=\bigcup_{U\in\mathscr{U}_0}V_U$, and let $Y=X\setminus V$. $A\subseteq V$, so $U\cap Y=(V_U\cup W_U)\cap Y$ for each $U\in\mathscr{U}$. Moreover, $V\in\tau$, so $Y$ is Lindelöf. Clearly $\{V_U\cup W_U:U\in\mathscr{U}\}$ is a $\tau$-open cover of $Y$, so there is a countable $\mathscr{U}_1\subseteq\mathscr{U}$ such that $Y\subseteq\bigcup_{U\in\mathscr{U}_1}(V_U\cup W_U)$ and hence $Y\subseteq\bigcup\mathscr{U}_1$. Then $\mathscr{U}_0\cup\mathscr{U}_1$ is a countable subset of $\mathscr{U}$ covering $V\cup Y=X$, and $X$ is $\tau(X\setminus A)$-Lindelöf.


Now suppose that $\langle X,\tau\rangle$ and $A\subseteq X$ are sequentially compact, and let $\sigma=\langle x_n:n\in\omega\rangle$ be a sequence in $X$. Since any constant sequence is convergent in any topology, we may assume that $x_m\ne x_n$ whenever $m\ne n$. Let $N=\{n\in\omega:x_n\in A\}$, and suppose that $N$ is infinite. $A$ is $\tau$-sequentially compact, so $\langle x_n:n\in N\rangle$ has a subsequence $\sigma'$ that $\tau$-converges to some $x\in A$. But $\tau$ and $\tau(X\setminus A)$ generate the same relative topology on $A$, so $\sigma'$ is a subsequence of $\sigma$ that $\tau(X\setminus A)$-converges to $x$.

If, on the other hand, $N$ is finite, we may assume that $N=\varnothing$, i.e., that $\sigma$ lies entirely in $X\setminus A$. $\langle X,\tau\rangle$ is sequentially compact, so we may assume (by passing to a subsequence if necessary) that $\sigma$ is $\tau$-convergent, say to $x$. Suppose that $x\in X\setminus A$; $\tau$ and $\tau(X\setminus A)$ generate the same relative topology on $X\setminus A$, so $\sigma$ converges to $x$ in $\tau(X\setminus A)$. Finally, suppose that $x\in A$, and with notation as before let $U=V_U\cup(W_U\setminus A)$ be a $\tau(X\setminus A)$-open nbhd of $x$. Then $x\in V_U\in\tau$, so there is an $m\in\omega$ such that $x_n\in V_U\subseteq U$ whenever $n\ge m$, and again we see that $\sigma$ converges to $x$ in $\tau(X\setminus A)$.

In all cases, therefore, a sequence in $X$ has a $\tau(X\setminus A)$-convergent subsequence, and $\langle X,\tau(X\setminus A)\rangle$ is therefore sequentially compact.

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