Question

Let X={a,b,c,d} with the topology T={ ϕ,{a},{a,b},{a,b,c,d}}.Find close set.

Answer 1

Answer on Question #70374, Math / Topology

Let $X = \{a, b, c, d\}$ with the topology $T = \{\varnothing, \{a\}, \{a, b\}, \{a, b, c, d\}\}$. Find close set.

**Solution.** Note that a subset $A$ of a topological space $X$ is closed, if $X \setminus A$ is an open subset of $X$. If $T$ is a topology of $X$, then $A$ set $A$ is open if and only if $A \in T$.

So, since $T = \{\varnothing, \{a\}, \{a, b\}, \{a, b, c, d\}\}$, then $X \setminus \varnothing = X$, $X \setminus \{a\} = \{b, c, d\}$, $X \setminus \{a, b\} = \{c, d\}$, $X \setminus \{a, b, c, d\} = \varnothing$.

Hence, subsets $\varnothing$, $X$, $\{c, d\}$ and $\{b, c, d\}$ are closed in $X$ and only they.

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