Quick proof that $\omega+1$ is not discrete space

Question

I asked a similar question yesterday but was stuck at this point.

Let $\omega+1$ be defined as:

$$\omega+1 = \mathbb{N} \cup \{\omega\}$$

Where $\omega \geq n, \forall n \in \mathbb{N}$, where $\geq$ is the usual ordering relationship

I wish to show that $\omega+1$ is not discrete (singletons are not open) with the order topology.

Since $\omega+1$ is a linear order, it is generated by the basis $$\mathcal{B} = \{(a,b)|a<b \in \omega+1\} \cup \{\varnothing\}$$

Then easily each singleton for $\mathbb{N}$ is open, since we can take, $\forall n \in \mathbb{N}$, $(n-1,n+1)= n$

The point that cause trouble is $\omega$. What is a good way to prove that $\{\omega\}$ is not open?

My attempt is here:

If $\{\omega\}$ was open, then $\{\omega\}^c$ is closed. However, $\{\omega\}^c = \mathbb{N}$ is open, since $\mathbb{N} = \bigcup_{n \in \mathbb{N}} \{n\}$ which is a countable union of open sets (we have shown singletons are open in above). Therefore $\{\omega\}$ is in fact closed.

Is this good enough?

Answer 1

Every open set contains a basic open set. Does $\{\omega\}$ contain an open interval?

Answer 2

For a linear order $<$ on a set $S$ we define an open interval as one of the following types of subset of $S$ :

$$(1)...S.$$ $$(2)...\emptyset.$$ $$(3)... (\leftarrow,x)=\{y\in S: y<x \} \text { for any } x\in S.$$ $$(4)... (x,\rightarrow)=\{y\in S:x<y\} \text { for any } x\in S.$$ $$(5)...(x,y)=\{z\in S: x<z<y\} \text { for any } x,y \in S.$$ The order topology $T$ on $S$ is the topology generated by the base of open intervals. So if $x\in U\in T$ then $x\in I\subset U$ for some open interval $I.$

In the case $S=N\cup \{\omega\},$ if $I$ is any open interval containing $\omega$ then $I= S\;$ (Type (1) ), or $I= (n,\rightarrow)= \{m\in N :m>n\}\cup \{\omega\}$ for some $n\in N\;$ (Type (4)). So every open interval containing $\omega,$ and hence every open set containing $\omega,$ contains members of $N .$ So $\{\omega\}$ is not open.