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Given $X = \{a, b, y_1, y_2, \ldots\}$, where $a$, $b$, $y_i$ are points.

The topology on $X$ is $\tau = \{X\} \cup T_f\cup T_d$, where $T_f$ is the cofinite topology on $X$ and $T_d$ is the discrete topology on $Y = \{y_1, y_2, \ldots\}$.

How do I show that each component of $X = \{a, b, y_1, y_2,\ldots\}$ is Hausdorff, but $X$ is not Hausdorff?


I'm pretty sure I'm not doing this right, but if this were an exam question and I'm desperate to write something down:

To show that $X$ is not Hausdorff, I need to show that there exists two nonempty sets in $T$ that intersect. Since $T_d\subseteq T$ and $T_d$ contains all subsets of $Y$, then obviously we can find two subsets of $Y$ in $T$ that intersect.

To show that every component of $X$ is Hausdorff, since $a$, $b$, $y_1$, etc are points, the singleton sets of $X$ are also the components, which are Hausdorff.

It feels kind of stupid :( Help please?

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    That's not what Hausdorff means.2012-03-19
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    First show that the components indeed are all the singletons, as you posit. This needs proof. This trivializes the second part. Then show that *all* neighbourhoods of $a$ and $b$ intersect.2012-03-19
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    @Andrea We say that $X$ is Hausdorff iff given any two points $x,y \in X$ such that $x \neq y$, **there exist disjoint neighbourhoods** $U$ and $V$ about $x$ and $y$ respectively. So a space $X$ is **not** Hausdorff if there exist points $x,y \in X$ with $x \neq y$ such that given ***any*** two neighbourhoods $U$ and $V$ about $x$ and $y$ respectively, $U \cap V \neq \emptyset$.2012-04-20

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