# Dense subset of a $T_1$ space

let X be a $T_1$ space without any isolated point and Y be dense subset of X. Show that for any open set U (non empty) in X, U intersection Y is infinite.

• What have you done?
– user87543
Dec 23, 2013 at 15:52
• Suppose not, then $U\cap Y$ is finite. Now what do you know about finite sets in a $T_1$ space? Dec 23, 2013 at 15:59
• Can you help , I am seeing that since y is dense in X so U intersection Y is non empty. I am not getting any clue. Give me a hint... Dec 23, 2013 at 16:01

Let $U$ be an open subset of $X$ such that $U\cap Y=${$x_1,x_2,...,x_n$}. Then $(X-${$x_1,x_2,...,x_n$})$\cap U\cap Y=\emptyset$$(*). Because {x_1,x_2,...,x_n} is closed in X (T_1 space) ,then (X-{x_1,x_2,...,x_n})\cap U is open (U is dense) and because Y is dense from the (*) we have contradiction. If not, assume that U\cap Y=\{x_1,\ldots,x_n\}. As X is T_1 space, then V_i=X\smallsetminus\{x_i\}, i=1,\ldots,n, are open, and also dense, as none of the x_1,\ldots,x_n are isolated points. It is not hard to show that the open set V=U_1\cap\cdots\cap U_n is also dense in X, and thus U\cap V is open and non-empty. Thus U\cap V \cap Y should be non-empty. Contradiction. if \mathfrak{T} is the collection of open sets, then the density of Y means that:$$\forall x \in X \; \forall U \in \mathfrak{T}. x \in U \rightarrow \exists y_1 \in Y \cap U$$choose a U_1 containing x. then by T1$$ \exists U_2 \in \mathfrak{T}. x \in U_2, y_1 \not \in U_2$$choose$y_2 \in U_2\$ (density again) etc.