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Math 344-1,2: Kursobjekt

Aaron Greicius

Section 1.5 Limit points and the Hausdorff property

Definition 1.5.1. Neighborhood of element or set.

Let \(X\) be a topological space. An open neighborhood of an element \(x\in X\) is an open set \(U\) that contains \(x\text{;}\) an open neighborhood of a subset \(Y\subseteq X\) is an open set \(U\) such that \(Y\subseteq U\text{.}\)
More generally, a (not necessarily open) neighborhood of a point \(x\in X\) or subset \(A\subseteq X\) is a subset \(K\) whose interior \(K^\circ\) contains \(x\) or \(A\text{:}\) equivalently, \(K\) contains an open set \(U\) containing \(x\) or \(A\text{.}\)

Definition 1.5.2. Limit point of a set.

Let \(A\) be a subset of the topological space \(X\text{.}\) An element \(x\in X\) is a limit point of \(A\) if any open neighborhood of \(x\) intersects \(A\) in a point distinct from \(x\text{:}\) i.e. for any open neighborhood \(U\) of \(x\text{,}\) we have \((U-\{x\})\cap A\ne \emptyset\text{.}\)

Example 1.5.3. Limit points in \(\R\).

Let \(X=\R\) with the standard topology. Determine the set \(L\) of all limit points for the following sets.
  1. \(\displaystyle (0,1)\)
  2. \(\displaystyle \Z\)
  3. \(\displaystyle K=\{1/n\colon n\in \Z_+\}\)
  4. \(\displaystyle \Q\)
Solution.
  1. We have $L=[0,1]$ in this case. For any \(x\in [0,1]\) and any metric ball \(B_\epsilon(x)\text{,}\) it is clear that $B_\epsilon(x)$ contains infinitely many elements of $[0,1]$. Thus $[0,1]\subseteq L$. Furthermore, since $[0,1]$ is closed, for any $x\notin [0,1]$ there is an open set $U$ containing $x$ such that $U\cap [0,1]=\emptyset$. Thus $L\subseteq [0,1]$.
  2. We have \(L=\emptyset\) in this. Indeed it is easy to see that for any $x\in \R$ we can find an open ball \(B_\epsilon(x)\) such that \(B_\epsilon(x)\cap \Z=\{x\}\) or \(B_\epsilon(x)\cap \Z=\emptyset\text{.}\)
  3. It is easy to see that $L=\{0\}$ in this case.
  4. In this case we have \(L=\R\text{:}\) for any \(x\in \R\) and any open ball \(B_\epsilon(x)\text{,}\) there are infinitely many elements of \(B_\epsilon(x)\cap \Q\text{.}\)

Proof.

We will show that \(\overline{A}\subseteq A\cup A'\) and \(A\cup A'\subseteq \overline{A}\text{.}\)

Inclusion \overline{A}\subseteq A\cup A’.

Assume \(x\in \overline{A}\) and \(x\notin A\text{.}\) Since \(x\in \overline{A}\text{,}\) for any open \(U\) containing \(x\text{,}\) there is any element \(y\in U\cap A\text{.}\) Since \(x\notin A\text{,}\) we must have \(x\ne y\text{.}\) Thus \(U-\{x\}\cap A=\emptyset\) and we see that \(x\in A'\text{.}\)

Inclusion \overline{A}\supseteq A\cup A’.

If \(x\in A\cup A'\text{,}\) then for all open sets \(U\) containing \(x\text{,}\) we have \(U\cap A\ne\emptyset\text{.}\) Thus \(x\in \overline{A}\text{.}\)

Proof.

Throughout we let \(A'\) be the set of limit points of \(A\text{.}\)

Implication: \((1)\implies (2)\).

If \(A\) is closed, then by Theorem 1.5.4 we have
\begin{equation*} A=\overline{A}=A\cup A'\text{.} \end{equation*}
It follows that \(A'\subseteq A\text{,}\) as desired.

Implication: \((2)\implies (3)\).

If \(A'\subseteq A\text{,}\) then again by Theorem 1.5.4 we have
\begin{equation*} \overline{A}=A\cup A'=A\text{.} \end{equation*}

Implication: \((3)\implies (1)\).

The implication is obvious since \(\overline{A}\) is closed.

Definition 1.5.6. Hausdorff property.

A topological space \(X\) is Hausdorff if for any two distinct elements \(x, y\in X\) (i.e., \(x\ne y\)) there are open neighborhoods \(U_x, U_y\) of \(x\) and \(y\text{,}\) respectively, that are disjoint (i.e., \(U_x\cap U_y=\emptyset\)).

Proof.

  1. Fix any \(x\in X\text{.}\) For any \(y\ne x\text{,}\) there exist disjoint open sets \(U, V\) containing \(x\) and \(y\text{,}\) respectively. In particular, we have \(V\cap \{x\}=\emptyset\text{.}\) It follows that \(y\notin \overline{\{x\}}\text{.}\) and thus that \(\overline{\{x\}}=\{x\}\text{.}\) Thus \(\{x\}\) is closed.
  2. This follows easily since a finite union of closed sets is closed.

Definition 1.5.8. The \(T_1\)-axiom.

A topological space is \(T_1\) (or is a \(T_1\)-space, or satisfies the \(T_1\)-axiom) if all finite subsets of \(X\) are closed.

Remark 1.5.9. \(T_1\) as separation axiom.

The \(T_1\) is called a separation axiom, or Trennungsaxiom in German: hence the ‘T’. To see more explicitly why this is called a separation axiom, note that (a) \(T_1\) is equivalent to the property that any singleton is closed, and (b) this in turn is equivalent to saying for any two elements \(x\ne y\) there is an open neighborhood \(U\) of \(y\) that does not contain \(x\text{;}\) we think of this \(U\) as separating \(x\) and \(y\text{.}\)
We will say more about separation axioms later. For now, I’ll mention that the Hausdorff property is itself a separation axiom, called the \(T_2\)-axiom.

Proof.

  1. This follows immediately from Theorem 1.5.7.
  2. We provide a counterexample to the claim that \(T_1\) implies Hausdorff. Let \(X\) be any infinite set endowed with the cofinite topology. The closed sets of \(X\) are precisely the finite sets and \(X\) itself, as we saw in Example 1.4.7. Thus \(X\) is \(T_1\text{.}\) On the other hand, \(X\) is not Hausdorff. Indeed take any two distinct elements \(x, y\in X\text{.}\) Let \(U_x\) and \(U_y\) be any pair of open neighborhoods of \(x\) and \(y\text{,}\) respectively. We will show that \(U_x\cap U_y\ne \emptyset\text{.}\) If either of the open neighborhoods is equal to \(X\) itself, the claim is obvious. Otherwise \(C=U_x^c\) is a finite set that does not contain \(x\) and \(D=U_y^c\) is a finite set that does not contain \(y\text{.}\) Since \(X\) is infinite, \(X\ne C\cup D\text{,}\) and hence
    \begin{equation*} U_x\cap U_y=C^c\cap D^c=(C\cup D)^c\ne \emptyset\text{,} \end{equation*}
    as desired.

Example 1.5.11. Metric spaces are Hausdorff.

  1. Show that any metric space is Hausdorff.
  2. Show that any infinite set endowed with the cofinite topology is not a metric space.
Solution.
  1. Let \((X,d)\) be a metric space. Given any \(x\ne y\in X\text{,}\) we have \(d(x,y)=r\ne 0\text{.}\) The open balls \(B_{r/2}(x), B_{r/2}(y)\) are disjoint open neighrborhoods of \(x\) and \(y\text{,}\) respectively.
  2. Let \(X\) be an infinite set with the cofinite topology. We will show that \(X\) is not Hausdorff and hence not a metric space. In fact, we will show that any two nonempty open sets in \(X\) intersect nontrivially. Indeed, given nonempty open sets \(U_1, U_2\text{,}\) we have by definition \(U_1=X-S_1\text{,}\) \(U_2=X-S_2\) for finite sets \(S_1, S_2\text{.}\) Their intersection is the open set
    \begin{equation*} U_1\cap U_2=(X-S_1)\cap (X-S_2)=X-(S_1\cup S_2). \end{equation*}
    Since \(X\) is infinite and the \(S_i\) are finite, we have \(X-(S_1\cup S_2)\ne\emptyset\text{,}\) and thus \(U_1\cap U_2\ne \emptyset\text{.}\)
The Zariski topology on \(\R^n\) is an important example of a non-Hausdorff topology. This topology plays an important role in algebraic geometry.

Topological specimen 8. Zariski topology on \(\R^n\).

Let \(X=\R^n\text{,}\) and let \(\R[x_1,x_2,\dots, x_n]\) be the set of polynomials in the unknowns \(x_i\) with real coefficients. For any \(S\subseteq \R[x_1,x_2,\dots, x_n]\) define
\begin{equation*} Z(S)=\{(a_1,a_2,\dots, a_n)\in \R^n\colon f(a_1,a_2,\dots, a_n)=0 \text{ for all } f\in S\}\text{.} \end{equation*}
In other words, \(Z(S)\) is the set of points \(P=(a_1,a_2,\dots, a_n)\) in \(\R^n\) that are simultaneous zeros for all polynomials \(f\in S\text{.}\) The set
\begin{equation*} \mathcal{T}=\{U\colon U=\R^n-Z(S) \text{ for some } S\subseteq \R^n\} \end{equation*}
defines a topology on \(X\) called the Zariski topology. As you will show in homework (for \(n=2\)), this topology is \(T_1\) but not Hausdorff.

Proof.

Proved in homework for \(n=2\text{.}\) The general proof is not much different.

Definition 1.5.12. Convergent sequence.

Let \(X\) be a topological space. A sequence \((x_n)_{n=1}^{\infty}\) of elements of \(X\) converges to the element \(x\in X\text{,}\) denoted \(x_n\to x\text{,}\) if for any open neighborhood \(U\)of \(x\text{,}\) there is a positive integer \(N\) such that \(x_n\in U\) for all \(n\geq N\text{.}\) We say \((x_n)_{n=1}^\infty\) is a convergent sequence in this case, and call \(x\) the limit of the sequence.

Example 1.5.13. Limits in the cofinite topology.

Let \(X=\R\) endowed with the cofinite topology. Show that the sequence \((n)_{n=1}^\infty=(1,2,3,\dots)\) converges to all elements \(x\in \R\text{.}\) In particular, the limit of a convergent sequence need not be unique!
Solution.
Let \(x\in \R\) and let \(U\) be any open set containing \(x\text{.}\) We have \(U=\R-\{r_1,r_2,\dots, r_n\}\) for some \(r_i\in \R\text{,}\) \(1\leq i\leq n\text{.}\) Since this list is finite, we can pick \(N\) such that if \(n\geq N\text{,}\) then \(n\ne r_i\) for any \(1\leq i\leq n\text{,}\) and hence that \(n\in U\text{,}\) as desired.
Note: the same argument shows that any sequence containing infinitely many distinct elements is convergent, and in fact converges to all elements of \(X\text{.}\)

Proof.

Assume \(x_n\rightarrow x\) and let \(y\ne x\text{.}\) Let \(U, V\) be disjoint open sets containing \(x, y\text{,}\) respectively. By convergence, there is an \(N\geq 1\) such that if \(n\geq N\text{,}\) then \(x_n\in U\text{.}\) It follows that \(x_n\notin V\) for all \(n\geq N\) and hence that \((x_n)\) does not converge to \(y\text{.}\)

Proof.

The reverse implication \((\impliedby)\) is obvious.
For the forward implication \((\implies)\) we will show the contrapositive. To this end, suppose there is an open neighborhood \(U\) of \(x\) such that \(U\cap A\) is finite. Since finite sets are closed in \(X\text{,}\) given any subset \(B\subseteq U\cap A\text{,}\) we have
\begin{equation*} U-B=U\cap B^c \end{equation*}
is open. Taking \(B=U\cap A-\{x\}\text{,}\) we see that \(V=U-B\) is open, contains \(x\text{,}\) and satisfies \(V\cap A\subseteq \{x\}\text{.}\) It follows that \(x\) is not a limit of point of \(A\text{,}\) as desired.
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