- #1
mathmari
Gold Member
MHB
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Hey! :giggle:
We consider the set $X=\mathbb{R}\cup \{\star\}$, i.e. $X$ consists of $\mathbb{R}$ and an additional point $\star$.
We say that $U\subset X$ is open if:
(a) For each point $x\in U\cap \mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset U$.
(b) If $\star \in U$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset U$.
Show that this defines a topology in $X$.
We have to show that $X$ and $\emptyset$ are open, the union of two open sets is open and the intersection of two open sets is open.
First we show that $X$ is open :
(a) For each point $x\in X\cap \mathbb{R}=\mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset X=\mathbb{R}\cup \{\star\}$. This is true since every neighboorhood of $x$ is contained.
(b) If $\star \in X$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset X=\mathbb{R}\cup \{\star\}$. Thisis true since the union of the intervals is a subspace of the real line.
Is that correct?
The emptyset is per definition open, or not? We don’t have to apply the given definition, do we? Let $M_1$ and $M_2$ be two open sets. We consider the union $M_1\cup M_2$. For each point $x\in M_1\cup M_2$ it is either $x\in M_1$ or $x\in M_2$ (or both) so statement (a) follows from the fact that $M_1$ and/or $M_2$ are open. The same holds also for statement (b).
Let $M_1$ and $M_2$ be two open sets. We consider the intersection $M_1\cap M_2$. For each point $x\in M_1\cap M_2$ it is $x\in M_1$ and $x\in M_2$ so statement (a) follows from the fact that $M_1$ and $M_2$ are open. The same holds also for statement (b). Therefore we get that the above defines a topology in $X$.
Is that correct and complete? :unsure:
We consider the set $X=\mathbb{R}\cup \{\star\}$, i.e. $X$ consists of $\mathbb{R}$ and an additional point $\star$.
We say that $U\subset X$ is open if:
(a) For each point $x\in U\cap \mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset U$.
(b) If $\star \in U$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset U$.
Show that this defines a topology in $X$.
We have to show that $X$ and $\emptyset$ are open, the union of two open sets is open and the intersection of two open sets is open.
First we show that $X$ is open :
(a) For each point $x\in X\cap \mathbb{R}=\mathbb{R}$ there exists an $\epsilon>0$ such that $(x-\epsilon, x+\epsilon)\subset X=\mathbb{R}\cup \{\star\}$. This is true since every neighboorhood of $x$ is contained.
(b) If $\star \in X$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset X=\mathbb{R}\cup \{\star\}$. Thisis true since the union of the intervals is a subspace of the real line.
Is that correct?
The emptyset is per definition open, or not? We don’t have to apply the given definition, do we? Let $M_1$ and $M_2$ be two open sets. We consider the union $M_1\cup M_2$. For each point $x\in M_1\cup M_2$ it is either $x\in M_1$ or $x\in M_2$ (or both) so statement (a) follows from the fact that $M_1$ and/or $M_2$ are open. The same holds also for statement (b).
Let $M_1$ and $M_2$ be two open sets. We consider the intersection $M_1\cap M_2$. For each point $x\in M_1\cap M_2$ it is $x\in M_1$ and $x\in M_2$ so statement (a) follows from the fact that $M_1$ and $M_2$ are open. The same holds also for statement (b). Therefore we get that the above defines a topology in $X$.
Is that correct and complete? :unsure:
