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16. The Subspace Topology.

Defn:  Let $(X, {\cal T})$ be a topological space, $Y \subset X$.  
Then the {\bf subspace topology} on $Y$ is the set 
\vskip 5pt
 \centerline{${\cal T}_Y = \{U \cap Y ~|~ U \in {\cal T} \}$}

\u
 $(Y, {\cal T}_Y)$ is a {\bf subspace} of $X$.

Lemma 16.1:  If ${\cal B}$ is a basis for the topology of $X$, then the
set 
\vskip 5pt
\centerline{${\cal B}_Y = \{B \cap Y ~|~ B \in {\cal B} \}$} 
\u
is a basis for the subspace topology on $Y$.

Lemma 16.2:  Let $Y$ be a subspace of $X$.  If $U$ is open in $Y$ and
$Y$ is open in $X$, then $U$ is open in $X$.

Lemma 16.3:  If $A_j$ is a subspace of $X_j$, $j = 1, 2$, then the
product topology on $A_1 \times A_2$ is the same as the topology $A_1
\times A_2$ inherits as a subspace of $X_1 \times X_2$.

\eject
Note:  Suppose $Y \subset X$ where $X$ is an ordered set with the order
topology. The order topology on $Y$ need not be the same as the
subspace topology on $Y$

Ex 1:$(0, 1) \cup \{5\}

Defn:  Suppose $Y \subset X$ where $X$ is an ordered set.  $Y$ is {\bf convex}  if for all $a, b \in Y$ such that $a < b$, then $(a, b) \subset Y$

Ex. 1:  $(1, 2) \cup (3, 4) \subset R$.

Ex. 2:  $(1, 2) \cup (3, 4) \subset (1, 2) \cup (3, 9)$.

Lemma 16.4:  Let $X$ is an ordered set with the order topology.  Let $Y$ be a convex subset of $X$.  Then the order topology on $Y$ is the same as the subspace topology on $Y$.

HW p91:  1, 3 (prove your answer), 8
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