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I find it difficult to understand one of the properties and the definition of Hausdorff spaces.

The property states that, "If $X$ is Hausdorff then every finite set is closed. How is this possible keeping in mind that the definition states that for two distinct elements you can find an open neighborhood around them such that their intersection is empty. Thus, wouldn't the finite set be open because you can always find an open neighborhood around distinct elements of finite sets?

Can someone please explain?

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    $\begingroup$ An open set in $X$! It might not be contained in your finite set! $\endgroup$ – Cocopuffs Sep 24 '13 at 3:39
  • $\begingroup$ Well firstly a set can be both open and closed, such sets are called clopen. For instance in every topological space $X$, both $\varnothing$ and $X$ are clopen, and more generally every connected component is clopen. $\endgroup$ – James Sep 24 '13 at 3:43
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So I decided to expand a little to give a proof the property you mention. Firstly a finite union of closed sets is closed (because this is the intersection of their complements which is open). So it suffices to show that a single point is a closed set (as given this take any finite union of them to get that finite sets are closed).

To see this let $x\in X$, for each $y\neq x\in X$, pick open, disjoint $U_y,V_y$ so that $x\in U_y,y\in V_y$.

Then $\bigcup_{y\in X} V_y$ is a union of open sets which is open. For each $y$, $y\in V_y$ so it is in the union and $x\notin V_y$ for each $y$ and so $x$ is not in the union.

Hence $\bigcup_{y\in X} V_y=X\setminus\{x\}$ is open and so its complement $\{x\}$ is closed.

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  • $\begingroup$ The intersection of open sets is not necessarily open. $\endgroup$ – Ian Macalinao Feb 9 '17 at 19:59
  • $\begingroup$ @simplyianm Indeed no, but it is if there are only finitely many, which is all I claim. $\endgroup$ – James Feb 9 '17 at 20:07
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I believe you are confused about what an open set is : Loosely speaking, an open set is one whose points are all interior points - a point $x$ in a set $U$ is called an interior point of $U$ if there is some "space" around it inside $U$.

For instance, consider the closed unit ball $$ \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \} $$ In this set, any point inside the ball (ie. $x^2+y^2 < 1$) is an interior point, but any point on the boundary (say, $(1,0)$) is not an interior point since there is no space to the right of $(1,0)$!

Thus, in a Hausdorff space $X$, if any $x \in X$, then $U = X\setminus \{x\}$ is open, since, for any point $y \in U$, there is some space around $y$ which does not touch $x$, and is thus contained in $U$.

With this in mind, try proving that the union of finitely many closed sets is closed (or, equivalently, the intersection of finitely many open sets is open).

Once you have this, go back to your question and try to prove it.

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  • $\begingroup$ Thank you so much for that explanation above. It would be so great to me if math books more frequently contain such clarification of the idea before the proof of theorems. $\endgroup$ – user286485 Sep 13 '16 at 16:48
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Let $F=\{x_1,...,x_n\}$ be a finite subset of a Hausdorff space. You want to show that $F^c$ is open.

Let $p \in F^c$, and for each $i$, there is some open set $\Pi_i$ containing $p$ that does not contain $x_i$. Let $\Pi = \cap_{i=1}^n \Pi_i$, then $\Pi$ is open, and contains no $x_i$, that is, $F \cap \Pi = \emptyset$. Hence $F^c$ is open.

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