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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.

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  • $\begingroup$ What have you done? $\endgroup$
    – user87543
    Dec 23, 2013 at 15:52
  • $\begingroup$ Suppose not, then $U\cap Y$ is finite. Now what do you know about finite sets in a $T_1$ space? $\endgroup$ Dec 23, 2013 at 15:59
  • $\begingroup$ 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... $\endgroup$
    – EuReka
    Dec 23, 2013 at 16:01

3 Answers 3

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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.

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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.

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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.

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