Topology!!!!

Definition 6.1   Let $X$ be a nonempty set. A collection $\tau \subset \mathcal{P}(X)$ is a topology if

(i)

$\emptyset, X \in \tau$

(ii)

If $G_1, G_2 \in \tau$ , then $G_1 \cap G_2 \in \tau$

(iii)

$\mathcal{G}\subset \tau$ , then $$\bigcup_{G \in \mathcal{G}} G \in \tau$$ .

The elements of $\tau$ are called the open sets.

The pair $(X, \tau)$ is a topological space.

Examples:

1. If $d$ is a metric on $X$ and $G \in \tau$ if and only if for all $x \in G$ , there exists $\epsilon > 0$ such that $B(x,\epsilon ) \subset G$ .
   Then $\tau$ is the metric topology on $X$ .
2. If $\tau = \{ \emptyset, X \}$ , then $\tau$ is a topology called the trivial topology.
3. If $\tau = \mathcal{P}(X)$ , then every set is open and $\tau$ is the discrete topology on$X$ .
   

Definition 6.2   A set is closed is $F^C \in \tau$ .

Definition 6.3   If $x \in X$ , a neighborhood of $x$ is any open set $G \in \tau$ with $x \in G$

Definition 6.4   If $(X, \tau)$ and $(Y, \sigma$ are topological spaces and $f: X \rightarrow Y$ is a function, then $f$ is continuous if whenever $H \in \sigma$ , $f^{-1}(H) \in \tau$ .

Remark: $f$ is continuous at $x \in X$ if and only if for all open subsets $H \in \sigma$ with $f(x) \in H$ , there exists a $G \in \tau$ such that $x \in G$ and $f(G) \subset H$ .

Example: Let $id$ be the identity map. Then,

$$id: (\mathbb{R},$$    metric topology $$) \rightarrow (\mathbb{R},$$    trivial $$)$$

is continuous, and

$$id: (\mathbb{R},$$    metric top. $$) \rightarrow (\mathbb{R},$$    discrete $$)$$

is not.

Definition 6.5   Let $\sigma$ and $\tau$ be topologies on $X$ . We say that $\sigma$ is weaker (or coarser) than $\tau$ if $\sigma \subset \tau$ .
We also say that $\tau$ is stronger (or finer) than $\sigma$ if $\sigma \subset \tau$ .

Definition 6.6   Let $(X, \tau)$ be a topological space and $E \subset X$ . The closure of $E$ is

$$\bar E := \bigcap_{F \text{ closed }, F \supset E} F$$

The interior $E^0$ of $E$ is

$$E^0 := \bigcap_{G \in \tau, G \subset E} G$$

Proposition 6.1   Let $X \neq \emptyset$ be any nonempty set, and let $\mathcal{C}\subset \mathcal{P}(X)$ . Then there is a weakest topology $\tau$ such that $\mathcal{C}\subset \tau$ .

Definition 6.7   Let $(X, \tau)$ be a topological space. A collection $\mathcal{B}\subset \tau$ is a base for $\tau$ if whenever $G \in \tau$ and $x \in G$ , then there exists $H \in \mathcal{B}$ such that

$$x \in H \subset G.$$

Monday, 3-28-2005:

Definition 6.8   A local base (or a neighborhood base) at $x \in X$ is a family $\mathcal{B}_x \subset \tau$ such that

(a)

$\forall G \in \mathcal{B}_x$ , $x \in G$ .

(b)

If $H \subset X$ is open and $x \in H$ , then there exists a $G \in \mathcal{B}_x$ such that $G \subset H$ .

Remark: Suppose $\mathcal{B}$ is a base for $\tau$ . Then, $G \in \tau \Leftrightarrow \forall x \in G$ , $\exists H \in \mathcal{B}$ such that $x \in H$ and $H \subset G$ .

Definition 6.9   A topological space is $1^{st}$ countable if every point $x \in X$ has a countable local base.

Definition 6.10   A topological space is $2^{nd}$ countable if $X$ has a countable base.

Examples:
Suppose $X$ is a metric space. Using balls of rational radius, we get that $X$ is $1^{st}$ countable. If, in addition, $X$ is separable, then $X$ is second countable.

Proposition 6.2   A family $\mathcal{B}$ of subsets of a set $X$ is a base for a topology $\tau$ on $X$ iff

(a

Each $x \in X$ belongs to some element of $\mathcal{B}$ (i.e., $$\bigcup_{H \in \mathcal{B}} H = X$$ ).

(b)

If $H_1, H_2 \in \mathcal{B}$ and $x \in H_1 \cap H_2$ , then $\exists H_3 \in \mathcal{B}$ such that $x \in H_3 \subset H_1 \cap H_2$ .

Definition 6.11   A collection $\mathcal{C}\subset \mathcal{P}(X)$ is a subbase for the topology $\tau$ if

$$\mathcal{B}:= \{ G_1 \cap \cdots \cap G_n : n \in \mathbb{N}, G_i \in \mathbb{C}, 1 \leq i \leq n \}$$

is a base for $\tau$ .

Example: Let $X = \mathbb{R}$ and $\mathcal{B}= \{ [a,b) : a < b \}$ .
Claim: $\mathcal{B}$ is a base for a topology $\tau$ -

(a)

holds as $$\bigcup_{H \in \mathcal{B}} H = \mathbb{R}$$ .

(b)

If $x \in H_1 \cap H_2$ , $H_i \in \mathcal{B}$ , then

$$x \in [a_1,b_1) \cap [a_2, b_2)$$

and

$$x \in [\max\{ a_1, a_2 \}, \min\{b_1, b_2\}) \in \mathcal{B}.$$

Hence, $\mathcal{B}$ is the base for a topology by our proposition.
This is called the ``half-open interval" topology. This topology is stronger than the usual Euclidean topology on $\mathbb{R}$ (it contains the Euclidean topology).
Notice

• The $\frac12$ -open interval topology is separable - the rational numbers are dense in this topology.
• This topology is first countable.
• This topology is not second countable. Suppose $\mathcal{A}$ is a base. Given $t \in \mathbb{R}$ , there exists a $G_t \in \mathcal{A}$ such that $G_t \subset [t,t+1)$ . Suppose $t < s$ . Then, $t \notin [s,s+1)$ , so $G_t \not\subset G_s$ .
  Thus, the map $t \rightarrow G_t$ is injective. Hence, $\mathcal{B}$ is uncountable, and this topology is not second countable. Thus, $\tau$ is not metrizable.
  

Friday, 4-1-2005

Example: On the Banach space $\mathcal{C}[0,1]$ , for $x \in [0,1]$ , let

$$\rho_x(f,g) = \vert f(x) - g(x)\vert.$$

Note that $\rho_x$ is a pseudometric (satisfies all the hypothesis of a metric except that $\rho_x(f,f)$ may be zero even if $f \neq 0$ .
Let $\mathcal{B}$ consist of all sets of the form

$$G_{x_1, \ldots, x_n, \epsilon }(f) := \{ g \in \mathcal{C}[0,1] : \rho_{x_i}(f,g) < \epsilon , i = 1, \ldots, n \}$$

Claim: $\mathcal{B}$ is the base for a topology.
Let

$$B_f = G_{x_1, \ldots, x_n, \epsilon _1}(f)$$

and

$$B_g = G_{y_1, \ldots, y_m, \epsilon _2}(g)$$

and suppose that $h \in B_f \cap B_g$ . Let

$$\{ z_1, \ldots, z_k\} = \{ x_1, \ldots, x_n \} \cup \{ y_1, \ldots, y_m \}$$

and

$$\epsilon = min \{ \epsilon _1, \epsilon _2 \}$$

Define

$$\delta = \max_{i,j} \{ \epsilon - \vert f(x_i) - h(x_i)\vert, \epsilon - \vert g(y_j) - h(y_j)\vert \}$$

Let

$$B_h = G_{z_1, \ldots, z_n, \delta} (h)$$

If $p \in B_h$ ,

$$\vert p(x_i) - f(x_i) \vert \leq \vert p(x_i) - h(x_i)\vert + \vert h(x_i) - f(x_i)\vert$$

$$\leq \epsilon _1 - \vert f(x) - h(x_i)\vert + \vert h(x_i) - f(x_i)\vert < \epsilon _1$$

So, $p \in B_f$ . A parallel argument will show that $p \in B_g$ . Hence, $B_h \subset B_f \cup B_g$ .
Now, if $f \in \mathcal{C}[0,1]$ , then certainly $\in B_{0, 1} (f)$ . Hence,

$$\bigcup_{B\in \mathcal{B}} B = \mathcal{C}[0,1]$$

Definition 6.12   Let $(X, \tau)$ be a topological space. Suppose $\{ x_n \}$ is a sequence in $X$ and $x \in X$ . We say that $x_n$ converges to $x$ if given any open set $G$ with $x \in G$ , there exists $N \in \mathbb{N}$ such that $x_n \in G$ for all $n \geq N$ .

Remark: A subbase for the topology on $\mathcal{C}[0,1]$ described above is the collection

$$\{ B_{x,\epsilon } : f \in \mathcal{C}[0,1], \epsilon > 0, x \in [0,1] \}$$

The topology on $\mathcal{C}[0,1]$ that we've just described is also called the topology of pointwise convergence. The reason for this name is that a sequence $f_n$ converges in $\mathcal{C}[0,1]$ in this topology if and only if $f_n$ converges pointwise.

Claim:
This topology $(\mathcal{C}[0,1], \mathcal{B})$ is not first countable. In particular, this means that $(\mathcal{C}[0,1], \mathcal{B})$ is not metrizable.

Proof. Let $f \in \mathcal{C}[0,1]$ and suppose $\{ B_i \}_{i=1}^\infty$ is a countable local base at $f$ . For each $i$ , we may find $\epsilon _i > 0$ and $x_{1,i}, x_{2,i}, \ldots, x_{n_i,i} \in [0,1]$ such that

$$G_{x_{1,i}, \ldots, x_{n_i,i},\epsilon _i}(f) \subset B_i$$

Let $t \in [0,1]$ be such that

$$t \notin \{ x_{1,1}, \ldots, x_{n_1,1}, x_{1,2}, \ldots, x_{n_2,2}, \ldots \}$$

Find a function $g_i \in \mathcal{C}[0,1]$ such that $\vert g_i(t) - f(t)\vert = 1$ and

$$g_i(x_{1,i} = f(x_1,i)$$

$$\vdots$$

$$g_i(x_{n_i},i) = f(x_{n_i},i)$$

Then, $G_i \notin G_{t,\frac12} (f)$ , but

$$g_i \in G_{x_{1,i}, \ldots, x_{n_i,i},\epsilon _i}(f) \subset B_i$$

Therefore, $g_i \in B_i \setminus G_{t,\frac12}(f)$ for all $i$ . Hence, there is no $i$ such that $f \subset B_i \subset G_{t,\frac12}(f)$ . Hence, $\{ B_i \}$ is not a local base at $f$ . $\qedsymbol$

Monday, 4-4-2005:
Motivation: Recall that a set $E$ in a metric space is closed if and only if whenever $(x_n) \subset E$ is a sequence in $E$ and $x_n \rightarrow x \in X$ , then $x \in E$ .

Example: Let $\tau$ be the topology of pointwise convergence in $\mathcal{C}[0,1]$ .
Let

$$E := \{ g \in [0,1] : 0 \leq g(x) \leq , \int_0^1 g(x) dx > 1 \}$$

Notice that if $g_n$ is a sequence in $E$ and $g_n \rightarrow g$ . Then, by the Dominated Convergence Theorem, $g \in E$ . Hence, from the metric space point of view, $E$ is closed.
However, this is not true in $\tau$ ; we claim that $0 \in \bar E$ . Let $G$ be a $\tau$ -open set such that $0 \in G$ . Then, there exists $\epsilon > 0$ and $x_1, \ldots, x_n \in [0,1]$ such that

$$0 \in G_{x_1,\ldots,x_n,\epsilon }(0) \subset G.$$

It is easy to see that we can find a function $f \in E$ such that $f \in G_{x_1,\ldots,x_n,\epsilon }(0)$ (take $f \equiv 2$ , then modify it so that it remains continuous and $f(x_i) = 0$ , making sure to keep the integral above 1).
Hence, $f \in G \cap E$ . Thus, $0 \in \bar E$ , and $E$ is not closed in this metric.
Hence, the set $E$ is sequentially closed, but not actually closed!

We would like to have an analogue of the above motivation, so something like it will hold true for all topological spaces.

Definition 6.13   Let $\Lambda$ be a nonempty set. A relation $\leq$ on $\Lambda$ is a direction if

(a)

$\forall \lambda \in \Lambda, \lambda \leq \lambda$ (reflexive).

(b)

If $\lambda_1, \lambda_2, \lambda_3 \in \Lambda$ and $\lambda_1 \leq \lambda_2$ , $\lambda_2 \leq \lambda_3$ , then $\lambda_1 \leq \lambda_3$ (transitive).

(c)

If $\lambda, \mu \in \Lambda$ , then $\exists \gamma \in \Lambda$ such that $\lambda \leq \gamma$ and $\mu \leq \gamma$ .

The pair $(\Lambda, \leq)$ is called a directed set

Examples:

(a)

$\mathbb{Z}, \mathbb{N}, \mathbb{R}, [0,\infty)$ are all directed sets with the usual $\leq$ .

(b)

Let $(X, \tau)$ be a topological space. Then, for $x \in X$ , let $\Lambda = \{ G \in \tau : x \in G \}$ . We define $G_1 \leq G_2$ if $G_2 \subset G_1$ . It is easy to see that the axioms for a directed set hold.

Definition 6.14   A net is a function $f: \Lambda \rightarrow X$ , where $\Lambda$ is a directed set.

For notation, we usually write $(x_\lambda)_{\lambda \in \Lambda}$ instead of $\{ f(\lambda) : \lambda \in \Lambda \}$ .

Examples: Any sequence is a net, where the domain is the natural numbers.

Example: Let $(X, \tau)$ be a topological space and let $\Lambda$ be the family of all neighborhoods of the given point $x \in X$ directed by reverse inclusion. For every $\lambda \in \Lambda$ , let $x_\lambda \in \lambda$ . Then, $(x_\lambda)_{\lambda \in \Lambda}$ is a net.

Example: Let $f:[0,1] \rightarrow \mathbb{R}$ be a function. Let $\mathcal{P}$ be the set of all partitions of the unit interval. Given a partition $P = \{ 0 = x_0 < \cdots < x_n = 1 \}$ , an $n$ -tuple $\tau = \{ c_1, \ldots, c_n )$ is compatible with $p$ if $c_j \in [x_{j-1},x_j]$ for all $j$ . Define

$$\Lambda = \{ (P, \tau) : P \in \mathcal{P}$$ and $\tau$ is compatible with $P$  $$\}$$

Define $(P_1,\tau_1) \leq (P_2,\tau_2)$ if $P_2$ refines $P$ . Then, $(\Lambda, \leq)$ is a directed set. Define a net in $\mathbb{R}$ by

$$x_{(P, \tau)} = \sum_{j=1}^n f(c_j) (x_j - x_{j-1})$$

We would like to define convergence on nets such that the net converges if and only if the function $f$ is Riemann Integrable.

Definition 6.15   Let $(X, \tau)$ be a topological space. Let $(x_\lambda)_{\lambda \in \Lambda}$ be a net. We say that $(x_\lambda)$ converges to $x \in X$ if, for all neighborhoods $G$ of $x$ , there is a $\lambda_0 \in \Lambda$ such that $(x)_\lambda \in G$ for all $\lambda \geq \lambda_0$ .

Wednesday, 4-6-2005:

Fact: Let $(X, \tau)$ be a topological space such that $E \subset X$ . Then,

$$\bar E = \{ x \in X :$$    whenever $G$ is a neighborhood of $x$ , $G \cap E \neq \emptyset$ $$\}$$

Proof. Let $x \in \bar E$ . Let $G$ be a neighborhood of $x$ . Then, $x \notin G^c$ , which is closed. Therefore, $G^c \not\supset E$ , and $G \cap E \neq \emptyset$ . $\qedsymbol$

Definition 6.16   Suppose $(X, \tau)$ is a topological space and $(x_\lambda)_{\lambda \in \Lambda}$ is a net in $X$ . We say that $(x_\lambda)$ is eventually in the set $E \subset X$ if there exists $\lambda_0 \in \Lambda$ such that $\lambda \geq \lambda_0$ implies $x_\lambda \in E$ .
We say that $(x_\lambda)$ is frequently in $E$ if for every $\lambda \in \Lambda$ , there exists $\mu \in \Lambda, \mu \geq \lambda$ and $x_\mu \in E$ .

Proposition 6.3   Let $(X, \tau)$ be a topological space. Let $E \subset X$ . Then,

$$\bar E = \{ x \in X : \exists$$    a net $$(x_\lambda)_{\lambda \in \Lambda}$$    s. t. $$x_\lambda \in E \forall \lambda$$    and $$x_\lambda \rightarrow x \}$$

Proof. Suppose $(x_\lambda)$ is a net in $E$ and $x_\lambda \rightarrow x$ .
Let $H$ be a neighborhood of $x$ . Then there exists some $\lambda_0 \in \Lambda$ such that $\lambda \geq \lambda_0$ implies $x_\lambda \in H$ . Then,

$$H \cap E \neq \emptyset$$

Hence, by the fact from the beginning of the class, we get that $x \in \bar E$ .
Conversely, suppose $x \in \bar E$ . Let $\Lambda$ be the set of all neighborhoods of $x$ directed by reverse inclusion. Then, for $\lambda \in \Lambda$ , we have $\lambda \cap E \neq \emptyset$ . So, pick $x_\lambda \in \lambda \cap E$ . Then, $(x_\lambda)_{\lambda \in \Lambda}$ is a net in $E$ .
If $H$ is a neighborhood of $x$ , then $H \in \Lambda$ . If $\lambda \geq H$ , then $\lambda \subset H$ . Hence, $(x_\lambda)$ is eventually in $H$ . Therefore, $x_\lambda \in x$ .
$\qedsymbol$

Proposition 6.4   Let $(X, \tau)$ and $(Y,\sigma)$ be topological spaces. Then a function $f: X \rightarrow Y$ is continuous if and only if whenever $(x_\lambda)_{\lambda \in \Lambda}$ is a convergent net $x_\lambda \rightarrow x$ we have $f(x_\lambda) \rightarrow f(x)$ .

Proof. Suppose $f$ is continuous and $x_\lambda \rightarrow x$ be a convergent net in $X$ . Let $H \subset Y$ be a neighborhood of $f(x) \in Y$ .
Then, $f^{-1}(H)$ is open in $X$ and $x \in f^{-1}(H)$ . Because the net $(x_\lambda)$ is eventually in $f^{-1}(H)$ , we get that $f(x_\lambda)$ is eventually in $H$ . Therefore, $f(x_\lambda) \rightarrow f(x)$ .
Conversely, suppose $f$ is not continuous. Then, there exists an open set $H \subset Y$ such that $f^{-1}(H)$ is not open. Hence, we can find $x \in f^{-1}(H)$ which is not an interior point of $f^{-1}(H)$ . Thus, if $\Lambda$ is the set of all neighborhoods of $x$ , again directed by reverse inclusion, we may find $x_\lambda \notin f^{-1}(H)$ and $x \in \lambda$ . Then, $(x_\lambda)$ is a net in $X$ and as before $x_\lambda \rightarrow x$ .
But $x_\lambda \notin f^{-1}(H)$ , so that $f(x_\lambda) \notin H$ for all $\lambda$ . This means that $f(x_\lambda) \not\rightarrow f(x)$ .
$\qedsymbol$

Example: Let $\mathcal{C}[0,1]$ be equipped with the topology of pointwise convergence.
For $f \in \mathcal{C}[0,1]$ , let

$$I(f) = \int_0^1 f(x) dx$$

Is $I$ continuous? No. Let

$$E = \{ f \in \mathcal{C}[0,1] : 0 \leq f \leq 2$$    and $$\int_0^1 f(x) dx \geq 1 \}.$$

Since $0 \in \bar E$ , there is a net $(f_\lambda) \subset E$ that converge to $E$ . However, $I(f_\lambda) \geq 1$ , so $I(f_\lambda) \not \rightarrow 0 = I(0)$ .
Hence, $I$ is not continuous.

Example: Let

$$B := \{ f \in \mathcal{C}[0,1] : \Vert f\Vert _\infty \leq 1 \}$$

It is easy to see that we can pick out an infinite sequence of functions in $B$ such that $\Vert f_i - f_j\Vert = 1$ for all $i \neq j$ . Hence, $B$ is not compact under the metric induced by $\Vert\cdot\Vert _\infty$ . However, we will later show that $B$ is $\tau$ compact!

Remark: Convergence can be used to define topologies. For example, let $X$ be a set and consider $\mathcal{P}(X)$ . If $(E_\lambda)$ is a net of sets (i.e., a net in $\mathcal{P}(X)$ , we can define the

$$\limsup E_\lambda = \bigcap_\lambda \bigcup_{\lambda \leq \mu} E_\mu$$

this is what we think it ``ought" to be. Then,

$$\liminf E_\lambda = \bigcup_\lambda \bigcap_{\lambda \leq \mu} E_\mu$$

We want to say that a sequence converges if and only if $\limsup E_\lambda = \liminf E_\lambda$ . This is the star topology.

Friday, 4-8-2005:

Examples of directed sets and nets

(a)

$\Lambda = \mathbb{R}\times \mathbb{R}$

$$(x_1,y_1) \leq (x_2,y_2) \Leftrightarrow x_1 \leq x_2$$    and $$y_1 \leq y_2$$

(b)

$\Lambda = \mathbb{R}$ , $\leq$ is usual order. Let $x_\lambda = e^{-\lambda}$ , $\lambda \in \mathbb{R}$ . Then,

$$\lim_{\lambda} x_\lambda = 0$$

Notice that this is a convergent net in $\mathbb{R}$ which is not bounded. On the other hand, if a sequence in $\mathbb{R}$ is convergent, then it is always bounded.

(c)

Let $\Lambda = \mathbb{R}$ ; define $x \leq_\Lambda y \Leftrightarrow y \leq x$ .
Define $x_\lambda = e^{-\lambda}$ as before. Then, this net is not convergent. Hence, even though the set is the same, the convergence is different depending on $\Lambda$ .

Definition 6.17   Let $(x_\lambda)_{\lambda \in \Lambda}$ be a net in the topological space $(X, \tau)$ . A point $x \in X$ is a cluster point for $(x_\lambda)_{\lambda \in \Lambda}$ if given any neighborhood $G$ of $x$ , $(x_\lambda)_{\lambda \in \Lambda}$ is frequently in $G$ .

Definition 6.18   Let $A$ be a directed set and let $\Lambda$ be a directed set. A function $\theta:A \rightarrow \Lambda$ is cofinal for $\Lambda$ if for all $\lambda_0 \in \Lambda$ , there exists an $a_0 \in A$ such that whenever $a \geq a_0$ , we have $\theta(a) \geq \lambda$ .

Definition 6.19   Let $(x_\lambda)_{\lambda \in \Lambda}$ be a net in $X$ . A subnet of $(x_\lambda)_{\lambda \in \Lambda}$ is the composition

$$A \rightarrow \Lambda \rightarrow X$$

of the net with a cofinal function $\theta:A \rightarrow \Lambda$ . We sometimes write $(x_{\lambda_A})_{a \in A}$ .

Remark: Suppose $(X, \tau)$ is a topological space and $(x_\lambda)$ is a net which converges, $x_\lambda \rightarrow x$ . Then every subnet of $(x_\lambda)$ converges to $x$ .

Example: Let $(e^{-t})_{t \in \mathbb{R}}$ (usual ordering) be the net described earlier. Then, $(e^{-n})_{n \in \mathbb{N}}$ is a subnet. Here, the map $\mathbb{N}\rightarrow \mathbb{R}$ is the inclusion map. However, $(e^{1-\frac1n})_{n \in \mathbb{N}}$ is not a subnet, as $n \mapsto 1 - \frac1n$ is not cofinal.

Example: Let $(e^{-n})_{n \in \mathbb{N}}$ be viewed as a net. Define $\theta: \mathbb{R}\rightarrow \mathbb{N}$ by

$$\theta(t) = \max \{ 0, [t] \}$$

Then, $(e^{-\theta(t)})_{t \in \mathbb{R}}$ is a subnet.
Moral: A subnet of a sequence need not be a sequence.

Proposition 6.5   Let $(x_\lambda)_{\lambda \in \Lambda}$ be a net in a topological space $X$ . Then $x \in X$ is a cluster point of $(x_\lambda)_{\lambda \in \Lambda}$ if and only if there exists a subnet of $(x_{\lambda_a})_{a \in A}$ which converges to $x$ .

Proof. Suppose $(x_{\lambda_a})_{a \in A}$ is a subnet such that $x_{\lambda_a} \rightarrow x$ . Then let $G$ be a neighborhood of $x$ . Let $\lambda_0 \in \Lambda$ . Since $a \mapsto \lambda_a$ is cofinal, there exists a $a_0 \in A$ such that whenever $a \geq a_0$ , $\lambda_a \geq \lambda_0$ .

Since $x_{\lambda_a} \rightarrow x$ , there exists an $a_1 \in A$ such that $x_{\lambda_a} \in G$ . Pick $a_2 \geq a_1$ and $a_2 \geq a_0$ . Then if $a \geq a_2$ , then $x_{\lambda_a} \in G$ and $\lambda_a \geq \lambda_0$ .

Hence, $(x_\lambda)$ is frequently in $G$ , so $x$ is a cluster point.

Monday, 4-11-2005:

We would now like to prove the converse. Let $\mathcal{N}$ be the collection of all neighborhoods of $x$ . Let $A = \mathcal{N}\times \Lambda$ and define $(G,\lambda) \leq_A (H,\mu)$ if and only if $\lambda \leq_\Lambda \mu$ and $H \subset G$ . Then $A$ becomes a directed set. Define $\theta:A \rightarrow \Lambda$ as follows:
Given $(N,\lambda) \in A$ , we know that since $x$ is a cluster point of $(x_\lambda)$ , $(x_\lambda)$ is frequently in $N$ . Hence, there exists $\mu \in \Lambda$ such that $x_\mu \in N$ and $\mu \geq \lambda$ .
Define $\theta(N,\lambda) = \mu$ where $\mu$ is chosen such that $\mu \geq \lambda$ and $x_\mu \in N$ . To see that $\theta:A \rightarrow \Lambda$ is a cofinal map, let $\lambda_0 \in \Lambda$ . Then if $(N,\lambda) \geq_A (X,\lambda_0)$ , we have $\theta(N, \lambda) \geq \lambda \geq \lambda_0$ . Hence, $\theta$ is a cofinal function. Hence,

$$(x_{\theta(N,\lambda)})_{(N,\lambda) \in A}$$

is a subnet of $(x_\lambda)_{\lambda \in \Lambda}$ .
Let $U$ be any neighborhood fo $x$ , and let $\lambda_0 \in \Lambda$ be fixed. If $(N,\lambda) \geq (U, \lambda_0)$ , then $x_{\theta(N,\lambda)} = x_\mu$ , where $\mu \in \Lambda$ satisfies $x_\mu \in N$ and $\mu \geq \lambda \geq \lambda_0$ .
So $x_\mu \in N \subset U$ and hence

$$x_{\theta(N,\lambda)} \in U, \forall (N,\lambda) \geq (U,\lambda_0)$$

Therefore,

$$(x_{\theta(N,\lambda)})_{(N,\lambda) \in A}$$

converges to $x$ . $\qedsymbol$