# Topology of $(-2,-1) \cup D[0,1] \cup \{1,2\}$

Definitions:

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If I understood the either, or, and, etc right, then $G=(-2,-1) \cup D[0,1] \cup \{1,2\}$ and thus

• 1.29 (a) $G$ is from left to right, an interval in $\mathbb R$, the unit disc, a point in $\mathbb R$ and then another point in $\mathbb R$.
• 1.29 (b) Interior: $D[0,1]$
• 1.29 (c) Boundary: $[-2,-1] \cup C[0,1] \cup \{1,2\} = [-2,-1) \cup C[0,1] \cup \{2\}$
• 1.29 (d) Isolated: $\{2\}$

1.30

Let $A=(-2,-1), B=D[0,1],C=\{1\},D=\{2\}$. I'm gonna group these into $G_1$ and $G_2$ separated s.t. $G_1 \subseteq G_3$ and $G_2 \subseteq G_4$

Ways:

1. $G_1=A \cup B \cup C, G_2 = D, G_3 = \{x < 1.5\}, G_4 = \{x > 1.5\}$
2. $G_1=A \cup D, G_2 = B \cup C, G_3 = \{x < -1\} \cup \{x > 1.5\}, G_4 = \{-1 < x < 1.5\}$
3. $G_1=A, G_2 = B \cup C \cup D, G_3 = \{x < -1\}, G_4 = \{x > -1\}$

Where have I gone wrong please?

• Did they really write $-2<z<-1$? I'd get another book.... Jul 29, 2018 at 11:56
• @LordSharktheUnknown Why? It's a perfectly valid property of $z \in \mathbb{C}$ to say that $z$ is real and $-2 < z < -1$. Jul 29, 2018 at 11:58
• Two places. First, note a boundary point of G need not be a member of G. Look again at -2. Second, none of your separations split B and C. That's correct, but suggests you should look again at the isolation of 1. Jul 29, 2018 at 12:15
• @DavidHartley Thanks! ^-^ I actually noticed those transcribed incorrectly. Post as answer?
– BCLC
Jul 29, 2018 at 12:17
• What does the notation $D[0,1]$ mean? Jul 29, 2018 at 13:40