MHB Proving Topology in X: A Look at Union & Intersection

[mathmari]

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:

 

[HOI]

No, you are completely missing the point! You are assuming that, because these are called "open sets", the properties of a toplogy must be true. This problem is asking you to prove that those properties hold for these particular sets. In particular, you have nowhere used the definitions of these sets.

 

[I like Serena]

(b) If $\star \in X$ then there is an $\epsilon>0$ such that $(-\epsilon , 0)\cup (0, \epsilon)\subset X=\mathbb{R}\cup \{\star\}$. Thisi s true since the union of the intervals is a subspace of the real line.

Hey mathmari!

I think it is all correct except that this sentence can be improved.

$\star$ is given as an element of $X$, which is what we should state rather than say "if $\star\in X$".
And the union of the intervals is "subset" instead of a "subspace".