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    Normality and compactness under a coarser topology

    P-T09A4

    Area
    Topology
    Topics
    Compactness, Point-Set Topology, Separation Axioms
    Seen at
    UNL
    Years
    2009
    Status
    draft

    Let \((X,\mathcal T)\) and \((X,\mathcal T')\) be topological spaces with \(\mathcal T\subseteq\mathcal T'\).

    1. If \((X,\mathcal T')\) is normal, must \((X,\mathcal T)\) also be normal?

    2. If \((X,\mathcal T')\) is compact, must \((X,\mathcal T)\) also be compact?

    Prove your answers in both parts.

    Solution

    1. The answer to part (a) is no: normality of the finer topology need not imply normality of the coarser topology.

    Let \[ X=\{a,b,c\} \] and define \[ \mathcal T = \{\varnothing,\{c\},\{a,c\},\{b,c\},X\}. \] This is a topology: unions of its members remain in the displayed list, and finite intersections do as well.

    Let \(\mathcal T'\) be the discrete topology on \(X\). Then \[ \mathcal T\subseteq\mathcal T'. \] The discrete space \((X,\mathcal T')\) is normal, since any two disjoint closed sets are themselves disjoint open neighborhoods of one another.

    In \((X,\mathcal T)\), the sets \[ \{a\} \qquad\text{and}\qquad \{b\} \] are closed because their complements \[ \{b,c\} \qquad\text{and}\qquad \{a,c\} \] belong to \(\mathcal T\). They are disjoint.

    Every open neighborhood of \(a\) contains \(c\): the only such neighborhoods are \(\{a,c\}\) and \(X\). Similarly every open neighborhood of \(b\) contains \(c\). Thus no open neighborhoods of \(\{a\}\) and \(\{b\}\) can be disjoint. Therefore \((X,\mathcal T)\) is not normal.

    2. The answer to part (b) is yes: if \((X,\mathcal T')\) is compact and \(\mathcal T\subseteq\mathcal T'\), then \((X,\mathcal T)\) is compact.

    Let \(\mathcal U\) be an arbitrary open cover of \(X\) by sets belonging to the coarser topology \(\mathcal T\). Because \[ \mathcal T\subseteq\mathcal T', \] every member of \(\mathcal U\) is also open in the finer topology \(\mathcal T'\). Hence \(\mathcal U\) is an open cover of the compact space \((X,\mathcal T')\). There are therefore finitely many members \[ U_1,\ldots,U_n\in\mathcal U \] with \[ X=U_1\cup\cdots\cup U_n. \] These same sets form a finite subcover in the coarser topology. Since \(\mathcal U\) was arbitrary, \((X,\mathcal T)\) is compact.

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