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$(X,\mathcal T)$ is a topological space and $A$ is dense in it and for each dense $B$, we have: $$B\subseteq A\to B=A$$ Is $X$ finite?

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up vote 4 down vote accepted

What if $\tau$ is the indiscrete topology, that is, $\tau=\{\varnothing,X\}$? Now every singleton is dense in $X$. Thus, if $A=\{x_0\}$ and $B$ is dense in $X$ with $B\subseteq A$, we have $B=A$.

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I like your's as well! and was quite surprised to see it (temporarily) disappear. – inactive... for now Apr 9 '13 at 11:23

Not necessarily. If $( X , \mathcal{T} )$ is a discrete space, then $X$ is the only dense subset. So any infinite discrete space will be a counterexample.

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+$1$ I like your counterexample! – Clayton Apr 9 '13 at 11:13

Any infinite topological space of Cantor-Bendixson rank one without condensation points is an example: the set of isolated points is a minimal dense subset.

For instance, take a sequence $(x_n)$ in a metric space with finitely many limit points $y_1,...,y_m$. Then $X= \{x_n \mid n \geq 0 \} \cup \{y_1,...,y_m\}$ and $A= \{x_n \mid n \geq 0\}$ is an example.

Other examples are ordinals less than $\omega_1$.

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