# Contents

## Statement

###### Proposition

Let $(X,\tau)$ be a topological space, and let $C \subset X$ be a closed subset, regarded as a topological subspace $(C,\tau_{sub})$. Then a subset $S \subset C$ is a closed subset of $(C,\tau_{sub})$ precisely if it is closed as a subset of $(X,\tau)$.

###### Proof

If $S \subset C$ is closed in $(C,\tau_{sub})$ this means equivalently that there is an open open subset $V \subset C$ in $(C, \tau_{sub})$ such that

$S = C \backslash V \,.$

But by the definition of the subspace topology, this means equivalently that there is a subset $U \subset X$ which is open in $(X,\tau)$ such that $V = U \cap C$. Hence the above is equivalent to the existence of an open subset $U \subset X$ such that

\begin{aligned} S & = C \backslash V \\ & = C \backslash (U \cap C) \\ & = C \backslash U \end{aligned} \,.

But now the condition that $C$ itself is a closed subset of $(X,\tau)$ means equivalently that there is an open subset $W \subset X$ with $C = X \backslash W$. Hence the above is equivalent to the existence of two open subsets $W,U \subset X$ such that

$S = (X \backslash W) \backslash U = X \backslash (W \cup U) \,.$

Since the union $W \cup U$ is again open, this implies that $S$ is closed in $(X,\tau)$.

Conversely, that $S \subset X$ is closed in $(X,\tau)$ means that there exists an open $T \subset X$ with $S = X \backslash T \subset X$. This means that $S = S \cap C = (X \backslash T) \cap C = C\backslash T = C \backslash (T \cap C)$, and since $T \cap C$ is open in $(C,\tau_{sub})$ by definition of the subspace topology, this means that $S \subset C$ is closed in $(C, \tau_{sub})$.

Created on May 15, 2017 at 12:15:04. See the history of this page for a list of all contributions to it.