J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. Supplementary Exercises: Nets
Definition
A poset \(J\) is directed iff, for all \(x,y \in J\), there exists \(z \in J\) such that \(x \leq z\) and \(y \leq z\).
Exercise 1
Show that the following are directed posets:
(a)
Any linearly ordered set.
Solution
For any \(x\), \(y\), we either have \(x \leq y\) or \(y \leq x\). In the first case, take \(z = y\). In the second case, take \(z = x\).
(b)
\(\mathcal{P} S\) for any set \(S\).
Solution
For any \(x,y \in \mathcal{P}S\) we have \(x \subseteq x \cup y\) and \(y \subseteq x \cup y\).
(c)
A set \(\mathcal{A}\) of subsets of \(S\) that is closed under finite intersection, partially ordered by \(\supseteq\).
Solution
For any \(X,Y \in \mathcal{A}\) we have \(X \supseteq X \cap Y\) and \(Y \supseteq X \cap Y\).
(d)
The set of all closed subsets of a topological space under \(\subseteq\).
Solution
For any \(C\), \(D\) closed we have \(C \cup D\) is closed and \(C, D \subseteq C \cup D\).
Exercise 2
A subset \(K\) of \(J\) is said to be cofinal in \(J\) if for each \(\alpha \in J\), there exists \(\beta \in K\) such that \(\alpha \leq \beta\). Show that if \(J\) is a directed poset and \(K\) is cofinal in \(J\), then \(K\) is directed.
Solution
- Let \(x,y \in K\)
- Pick \(z \in J\) such that \(x \leq z\) and \(y \leq z\).
- Pick \(t \in K\) such that \(z \leq t\).
- \(x \leq t\) and \(y \leq t\)
Exercise 3
Let \(X\) be a topological space. A net in \(X\) is a function \(f\) from a directed poset \(J\) into \(X\). If \(\alpha \in J\), we usually denote \(f(\alpha)\) by \(x_\alpha\). We denote the net \(f\) itself by the symbol \((x_\alpha)_{\alpha \in J}\), or merely by \((x_\alpha)\) if the index set is understood.
The net \((x_\alpha)\) is said to converge to the point \(x \in X\) (written \(x_\alpha \rightarrow x\)) if for each neighbourhood \(U\) of \(x\), there exists \(\alpha \in J\) such that
\[ \alpha \leq \beta \Rightarrow x_\beta \in U \]Show that these definitions reduce to familiar ones when \(J = \mathbb{Z}_+\).
Solution
Just an observation.
Exercise 4
Suppose that
\[ (x_\alpha)_{\alpha \in J} \rightarrow x \text{ in } X \text{ and } (y_\alpha)_{\alpha \in J} \rightarrow y \text{ in } Y \enspace . \]Show that \((x_\alpha, y_\alpha) \rightarrow (x,y)\) in \(X \times Y\).
Solution
- Let \(U\) be a neighbourhood of \((x,y)\).
- Pick neighbourhoods \(V\) of \(x\) and \(W\) of \(y\) such that \(V \times W \subseteq U\).
- Pick \(\alpha, \beta \in J\) such that \(\forall \gamma \geq \alpha. x_\gamma \in V\) and \(\forall \gamma \geq \beta. y_\gamma \in W\).
- Pick \(\delta \in J\) such that \(\alpha \leq \delta\) and \(\beta \leq \delta\).
- \(\forall \gamma \geq \delta. (x_\gamma, y_\gamma) \in V \times W \subseteq U\)
- \(\Box\)
Exercise 5
Show that if \(X\) is Hausdorff, a net in \(X\) converges to at most one point.
Solution
- Assume \(X\) is Hausdorff.
- Assume for a contradiction \((x_\alpha) \rightarrow l\) and \((x_\alpha) \rightarrow m\) with \(l \neq m\).
- Pick disjoint neighbourhoods \(U\) and \(V\) of \(l\) and \(m\).
- Pick \(\alpha, \beta \in J\) such that \(\forall \gamma \geq \alpha. x_\alpha \in U\) and \(\forall \gamma \geq \beta. x_\beta \in V\).
- Pick \(\delta \in J\) such that \(\alpha \leq \delta\) and \(\beta \leq \delta\).
- \(x_\delta \in U \cap V\)
- This is a contradiction.
- \(\Box\)
Exercise 6
Theorem. Let \(A \subseteq X\). Then \(x \in \overline{A}\) if and only if there is a net of points of \(A\) converging to \(x\).
Solution
- If \(x \in \overline{A}\) then there exists a net of points of \(A\) converging to \(x\).
- Assume \(x \in \overline{A}\).
- Let \(J\) be the poset of neighbourhoods of \(x\) under \(\supseteq\).
- \(J\) is directed.
- If \(U,V \in J\) then \(U \cap V \in J\) and \(U, V \supseteq U \cap V\).
- For \(U \in J\), pick \(a_U \in U \cap A\). Prove: \(a_U \rightarrow x\).
- Theorem 17.5
- Let \(U\) be any neighbourhood of \(x\).
- We have \(\forall V \subseteq U. a_V \in U\).
- If there exists a net of points of \(A\) converging to \(x\) then \(x \in \overline{A}\).
- For any neighbourhood \(U\) of \(x\), there exists \(\alpha\) such that \(x_\alpha \in U \cap A\).
- \(\Box\)
Exercise 7
Theorem. Let \(f : X \rightarrow Y\). Then \(f\) is continuous if and only if, for every convergent net \((x_\alpha)\) in \(X\), converging to \(x\), say, the net \((f(x_\alpha))\) converges to \(f(x)\).
Solution
- If \(f\) is continuous then, whenever \(x_\alpha \rightarrow x\), we have \(f(x_\alpha) \rightarrow f(x)\).
- Assume \(f\) is continuous.
- Let \(x_\alpha \rightarrow x\).
- Let \(V\) be a neighbourhood of \(f(x)\).
- \(f^{-1}(V)\) is a neighbourhood of \(x\).
- Pick \(\alpha \in J\) such that \(\forall \beta \geq \alpha. x_\beta \in f^{-1}(V)\).
- \(\forall \beta \geq \alpha. f(x_\beta) \in V\)
- If, whenever \(x_\alpha \rightarrow x\), we have \(f(x_\alpha) \rightarrow f(x)\), then \(f\) is continuous.
- \(\langle 2 \rangle 1\) Assume that, whenever \(x_\alpha \rightarrow x\), we have \(f(x_\alpha) \rightarrow f(x)\).
- Let \(A \subseteq X\). Prove: \(f(\overline{A}) \subseteq \overline{f(A)}\).
- Let \(x \in \overline{A}\). Prove: \(f(x) \in \overline{f(A)}\).
- Pick a net of points \((x_\alpha)\) in \(A\) that converges to \(x\).
- Exercise 6
- \(f(x_\alpha) \rightarrow f(x)\)
- \( \langle 2 \rangle 1 \)
- \(f(x) \in \overline{f(A)}\)
- Exercise 6
- \(\Box\)
Exercise 8
Let \(f : J \rightarrow X\) be a net in \(X\); let \(f(\alpha) = x_\alpha\). If \(K\) is a directed set and \(g : K \rightarrow J\) is a function such that
- \(i \leq j \Rightarrow g(i) \leq g(j)\)
- \(g(K)\) is cofinal in \(J\)
then the composite function \(f \circ g : K \rightarrow X\) is called a subnet of \((x_\alpha)\). Show that if the net \((x_\alpha)\) converges to \(x\), so does any subnet.
Solution
- Assume \(x_\alpha \rightarrow x\).
- Let \(g : K \rightarrow J\) be a function satisfying 1 and 2. Prove: \(x_{g(k)} \rightarrow x\).
- Let \(U\) be any neighbourhood of \(x\).
- Pick \(\alpha \in J\) such that \(\forall \beta \geq \alpha. x_\beta \in U\).
- Pick \(k \in K\) such that \(\alpha \leq g(k)\). Prove: \(\forall k' \geq k. x_{g(k')} \in U\).
- Let \(k' \geq k\).
- \(g(k') \geq g(k)\)
- \(g(k') \geq \alpha\)
- \(x_{g(k')} \in U\)
- \(\Box\)
Exercise 9
Let \((x_\alpha)_{\alpha \in J}\) be a net in \(X\). We say that \(x\) is an accumulation point of the net \((x_\alpha)\) if for each neighbourhood \(U\) of \(x\), the set \(\{\alpha \in J \mid x_\alpha \in U\}\) is cofinal in \(J\).
Lemma. The net \((x_\alpha)\) has the point \(x\) as an accumulation point if and only if some subnet of \((x_\alpha)\) converges to \(x\).
Solution
- If \(x\) is an accumulation point of \((x_\alpha)\) then there exists a subnet of \((x_\alpha)\) that converges to \(x\).
- Assume \(x\) is an accumulation point of \((x_\alpha)\).
- Let \(K = \{(\alpha, U) \mid \alpha \in J, U \text{ is a neighbourhood of } x \text{ that contains } x_\alpha\}\).
- Define \(\leq\) on \(K\) by \((\alpha, U) \leq \(\beta, V)\) iff \(\alpha \leq \beta\) and \(V \subseteq U\).
- \(K\) is a directed poset.
- Let \((\alpha, U), (\beta, V) \in K\)
- Pick \(\gamma \in J\) such that \(\alpha, \beta \leq \gamma\)
- Pick \(\delta \in J\) such that \(x_\delta \in U \cap V\) and \(\gamma \leq \delta\).
- Since \(x\) is an accumulation point. l>
- \((\alpha, U), (\beta, V) \leq (\delta, U \cap V)\)
- Define \(g : K \rightarrow J\) by \(g((\alpha, U)) = \alpha\).
- \((x_{g(\alpha,U)})\) is a subnet of \((x_\alpha)\).
- If \((\alpha, U) \leq (\beta, V)\) then \(\alpha \leq \beta\).
- \(g(K)\) is cofinal in \(J\).
- Let \(\alpha \in J\)
- We have \((\alpha, X) \in K\) and \(g(\alpha, X) = \alpha\)
- \(x_{g(\alpha,U)} \rightarrow x\)
- Let \(U\) be a neighbourhood of \(x\).
- Pick \(\alpha \in J\) such that \(x_\alpha \in U\)
- For all \((\beta,V) \geq \(\alpha, U)\) we have \(x_\beta \in U\)
- If there exists a subnet of \((x_\alpha)\) that converges to \(x\) then \(x\) is an accumulation point of \((x_\alpha)\).
- Assume \(x_{g(k)} \rightarrow x\)
- Let \(U\) be a neighbourhood of \(x\).
- Let \(\alpha \in J\)
- Pick \(k \in K\) such that \(\forall k' \geq k. x_{g(k')} \in U\)
- Pick \(\beta \in J\) such that \(g(k), \alpha \leq \beta\)
- Pick \(k'' \in K\) such that \(\beta \leq g(k'')\)
- Pick \(k_0 \in K\) such that \(k, k'' \leq k_0\)
- \(\forall k_1 \geq k_0. x_{g(k_1)} \in U\)
- \(\Box\)
- If \(X\) is compact then every net in \(X\) has a convergent subnet.
- Assume \(X\) is compact.
- Let \((x_\alpha)_{\alpha \in J}\) be a net in \(X\).
- For \(\alpha \in J\), let \(B_\alpha = \{x_\beta \mid \alpha \leq \beta\}\).
- \(\{\overline{B_\alpha} \mid \alpha \in J\}\) has the finite intersection property.
- Pick \(l \in \bigcap_{\alpha \in J} B_\alpha\).
- \(l\) is an accumulation point of \((x_\alpha)\).
- Let \(U\) be a neighbourhood of \(l\).
- Let \(\alpha \in J\).
- \(U\) intersects \(B_\alpha\)
- Since \(l \in \overline{B_\alpha}\).
- There exists \(\beta \geq \alpha\) such that \(x_\beta \in U\).
- Some subnet of \((x_\alpha)\) converges to \(l\).
- Exercise 9
- \((x_\alpha)\) has a convergent subnet.
- If every net in \(X\) has a convergent subnet then \(X\) is compact.
- Assume every net in \(X\) has a convergent subnet.
- Let \(\mathcal{A}\) be a set of closed sets in \(X\) with the finite intersection property.
- Let \(\mathcal{B}\) be the set of all finite intersections of elements of \(\mathcal{A}\) partially ordered by \(\supseteq\).
- For \(B \in \mathcal{B}\), pick \(x_B \in B\).
- Pick a convergent subnet \((x_{g(k)})_{k \in K}\) of \((x_B)_{B \in \mathcal{B}}\) that converges to \(l\), say. Prove: \(l \in \bigcap \mathcal{A}\).
- Let \(A \in \mathcal{A}\). Prove: \(l \in A\).
- Every neighbourhood of \(l\) intersects \(A\).
- Let \(U\) be a neighbourhood of \(l\).
- Pick \(k \in K\) such that \(\forall k' \geq k. x_{g(k')} \in U\)
- Pick \(k' \in K\) such that \(A \cap g(k) \supseteq g(k')\)
- \(x_{g(k')} \in A cap U\)
- \(l \in \overline{A}\)
- \(l \in A\)
- \(\Box\)
- Let \(x \in \overline{AB}\). Prove: \(x \in AB\).
- Pick a net \((a_\alpha b_\alpha)_{\alpha \in J}\) in \(AB\) that converges to \(x\).
- Exercise 6
- Pick a convergent subnet \((b_{g(k)})_{k \in K}\) of \((b_\alpha)\) that converges to \(b\), say.
- Exercise 10
- \((a_{g(k)}b_{g(k)},b_{g(k)}^{-1}) \rightarrow (x,b^{-1})\)
- Exercises 4 and 7
- \(a_{g(k)} = a_{g(k)} b_{g(k)} b_{g(k)}^{-1} \rightarrow xb^{-1}\)
- Exercise 7
- \(xb^{-1} \in A\)
- Exercise 6
- \(x \in AB\)
- \(\Box\)
Exercise 10
Theorem. \(X\) is compact if and only if every net in \(X\) has a convergent subnet.
Solution
Exercise 11
Corollary. Let \(G\) be a topological group. Let \(A, B \subseteq G\). If \(A\) is closed and \(B\) is compact then \(AB\) is closed.
Solution
Exercise 12
Check that the preceding exercises remain correct if condition (2) is omitted from the definition of directed set. Many mathematicians use the term "directed set" in this more general sense.
Solution
The above solutions are all valid with the alternative definition of directed set.
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