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I hope that someone could give me hint how to solve this problem.

Let $(X,\tau)$ be a topological space. Prove that $\tau$ is the finite-closed topology on $X$ if and only if (i) $(X,\tau)$ is a $T_1$ space, and (ii) every infinite subset of $X$ is dense in $X$.

where the finite-closed topology is defined as all finite subsets of $X$ and X itself are closed, and $T_1$ space is defined as all singleton of $X$ is closed.

What I have done for "if $\tau$ is the finite closed topology on $X$, then $(X,\tau)$ is a $T_1$ space, and every infinite subject of X is dense in X" is as follows

Since all finite subsets of $X$ is closed and all singletons of $X$ is finite, all singletons of $X$ must be closed, and hence the finite-closed topology is a $T_1$ space. Suppose some infinite subset, $S$, of $X$ is not dense in $X$, $S\cup S' \ne \emptyset$, where $S'$ is a set of $S$'s limit points. This means for some $x \notin S$, there exist at least one open set $U_x$ such that $x\in U_x$ and $U_x \cap S = \emptyset$. Then, $X\backslash U_x$ must be finite by definition of the finite-closed topology,....

Then I have no idea how to proceed, intuitively, I think if $U_x \cap S = \emptyset$, then $X\backslash U_x$ must be infinite subset of $X$. I am not eve sure this is correct intuition. Also I have no idea how to approach for the second part "if $(X,\tau)$ is a $T_1$ space and every infinite subset of $X$ is dense in $X$, then $\tau$ is the finite closed topology on $X$.

Please give me some hint,,,,

Thanks,

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If you have a subset $S\subseteq X$ that is not dense, then there exists an open subset $U\subseteq X$ such that $U\cap S = \varnothing$. Since $U$ is open, $X\setminus U$ is finite and hence $S\subseteq X\setminus U$ is finite.

Conversely, assume that $(X,\tau)$ is (i) $T_1$ and (ii) every infinite subset is dense. Then every finite set is closed by (i), and if you take any closed subset $A\subseteq X$, then $A$ is not dense (because $X\setminus A$ is open and $A\cap (X\setminus A) = \varnothing$) and thus $A$ is finite by (ii).

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