Closed Subset of a Complete Space is Complete

Last updated Nov 6, 2022

Let $(X, d)$ be a Complete Metric Space. Suppose $C \subset X$ is Closed. Then $(C, d {\big|}_{C \times C})$ is a Complete Metric Space.

TODO

• A Set is Closed in a Metric Space iff it contains all its Sequential Limits
• definition of Complete Metric Space

Let $(X, \leq)$ be a Complete Ordering equipped with the Order Topology. Suppose $C \subset X$ is Closed. Then $(C, \leq {\big|}_{C \times C})$ is a Complete Ordering.

If $C = \emptyset$ then this is vacuously true. Otherwise, assume $C \neq \emptyset$. Let $A \subset C$ be Nonempty be bounded from above. Since $X$ is a Complete Ordering we know $\sup\limits A$ exists. Since a Supremum is in the Closure of a Nonempty Set, we know $\sup\limits A \in \text{cl} A$. Because Closure is Monotonic, $\text{cl} A \subset C$ and $\sup\limits A \in C$. Therefore $(C, \leq {\big|}_{C \times C})$ is a Complete Ordering. $\blacksquare$

• Subspace Metric Space
• Empty Set
• Nonempty