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    Subnets of convergent nets converge

    E-2BNJ6

    Area
    Topology
    Topics
    Nets
    Status
    draft

    Let \(f: J \to X\) be a net in \(X\); let \(f(\alpha) = x_\alpha\). If \(K\) is a directed set and \(g: K \to J\) is a function such that

    1. \(i \preceq j \implies g(i) \preceq g(j)\),

    2. \(g(K)\) is cofinal in \(J\),

    then the composite function \(f \circ g: K \to X\) is called a subnet of \((x_\alpha)\). Show that if the net \((x_\alpha)\) converges to \(x\), so does any subnet.

    Solution

    Goal: Prove that every subnet \((y_k)_{k \in K} = (x_{g(k)})_{k \in K}\) of a convergent net \((x_\alpha)_{\alpha \in J} \to x\) in a topological space \(X\) also converges to \(x\).

    1. Target neighborhood:Let \(U\) be an arbitrary open neighborhood of \(x\) in \(X\).

    2. Convergence of the parent net \((x_\alpha)\):There exists an index \(\alpha_0 \in J\) such that for all \(\alpha \in J\):\[\alpha_0 \preceq \alpha \implies x_\alpha \in U.\]Proof: Follows directly from the definition of net convergence \((x_\alpha) \to x\).

    3. Cofinality of the subnet indexing map \(g\):There exists an element \(k_0 \in K\) such that \(\alpha_0 \preceq g(k_0)\).Proof: By condition (ii), \(g(K)\) is cofinal in \(J\). Applying cofinality to the element \(\alpha_0 \in J\) provides such a \(k_0 \in K\).

    4. Monotonicity and convergence verification on \(K\):For every \(k \in K\) with \(k_0 \preceq k\), \(y_k = x_{g(k)} \in U\).Proof:

    4.1. Let \(k \in K\) satisfy \(k_0 \preceq k\).

    4.2. By condition (i), the map \(g: K \to J\) is order-preserving, so \(g(k_0) \preceq g(k)\).

    4.3. By transitivity of the preorder \(\preceq\) on \(J\), \(\alpha_0 \preceq g(k_0)\) and \(g(k_0) \preceq g(k)\) imply:\[\alpha_0 \preceq g(k).\]

    4.4. By 2, since \(\alpha_0 \preceq g(k)\), the net value satisfies \(x_{g(k)} \in U\).

    4.5. Therefore \(y_k = (f \circ g)(k) = x_{g(k)} \in U\).

    5. Conclusion:For every open neighborhood \(U\) of \(x\), there exists \(k_0 \in K\) such that \(k_0 \preceq k \implies y_k \in U\). Hence the subnet \((y_k)_{k \in K}\) converges to \(x\). Q.E.D.