Showing a series converges to a point in the usual topology

Question

Show that the sequence $s_n = \frac{n}{n+1} $ converges to 1 in $(\mathbb{R}, \mathscr{U})$ but does not converge to 1 in $(\mathbb{R}, \mathscr{H})$

The definition I have for converges to $x$ is let $X$ be a topological space and let $x \in X$. A sequence $s$ in $X$ is said to converge to $x$ provided that, for any open set $U$ containing $X$, there exists a positive integer $N$ such that if $n>N $, then $s_n \in U$.

I'm struggling to apply the definition to prove both parts of the question.

Answer 1

So the first thing we observe is that $s_n$ is increasing and bounded above. That is $s_1<s_2<\dots<1.$

Now let $A$ be an open neighbourhood of $1$ in the usual topology. As open intervals are a basis for this, let $(a,b)\subset A$ with $a<1<b.$ then we solve $\frac x{x+1}=a$ to get $x = \frac a{1-a},$ and by the Archimedes axiom we have some $n>x$ an integer. By the properties above we have $$\forall m>n \quad b>1 > s_m >s_n>a.$$ Therefore we are done.

For the next part consider $A=[1,2),$ which is an open neighbourhood of 1 under the half open topology. Yet by properties above it contains no $s_i.$

Answer 2

In the usual topology, any nbhd of $1$ contains an interval $(a,b)$ with $1\in (a,b)$. $\frac n{n+1}=\frac1{1+\frac1n}$ can be made arbitrarily close to $1$ for sufficiently large $n$.

It doesn't converge in the half open interval topology, because the nbhd $[1,a)$ doesn't contain a point of the sequence. Note that the sequence increases...