Imposing the topology of open rays in $\Bbb R$ After having received Brian M. Scott's permission (see comments in the selected answer) I am integrating his suggestions with my own solutions to form a complete answer to the questions apperaing below. 

Let $\mathscr{T}$ be the collection of subsets of $\Bbb R$ consisting
  of $\emptyset, \Bbb R$ and the rays of the form $(r, \infty)$,
  where $r \in \Bbb R$.
$(a)$ Exhibit that this, indeed, is a topology on $\Bbb R$.

Proof: Given any finite number of open sets of the form above, $\exists$ a maximal $r_n$. The set $(r_n, \infty)$ is the intersection of these finite sets. Let $\{U_i\}_{i \in I}$ be an arbitrary family of open sets. Let $r_{\min} = \inf\{r_i\}$. Then $(r_{\min}, \infty)$ is the union of the $U_i$ and clearly it is of the desired form to be an element of the topology. If $r_{\min} = -\infty$ then the union of the $U_i$ is the entire $\Bbb R$.

$(b)$ Show that it fails to be a topology if $r \in \Bbb Q$.

I don't have the full answer to this. I think that there should be a problem in the union of arbitrarily many open sets. Any help with a counterexample will be very helpful.

Answer the following questions. Is $(\Bbb R, \mathscr{T})$:
$(c)$ $T_1$?

No. Let $x_1 \neq x_2 \in \Bbb R$ and assume without loss of generality that $x_1 < x_2$. Then any open set containing $x_1$ will be of the form $x_1 - \epsilon, \infty$ for some $\epsilon > 0$ and will encessarily contain $x_2$.

$(d)$ Hausdorff

No. Being Hausdorff (or $T_2$) would imply that it is $T_1$, a contradiction to (a).

$(e)$ metrizable

No. If there was a metric, the metric space would have to be Hausdorff, contradicting (b).

$(f)$ second - countable

I have no idea on how to go about this one.

$(g)$ compact

No. There exists no finite cover of this space. Suppose we are given a finite cover of this space. Then there exists a minimal $r$ such that $(r_{\min}, \infty)$ covers the entire space. Of ,course, this is only possible if $(r_{\min}, \infty) = \Bbb R$.

$(h)$ locally compact

Yes. We need to exhibit that any point has a compact neighbourhood. To this end, fix $x \in \Bbb R$. Let $(r, \infty)$ be a neighbourhood of $x$ and let $\{U_i\}_{i \in I}$ be an open cover. Then, there must exist $q \in \Bbb R$ such that $q \ge r$ so that $(q, \infty)$ is a set of the open cover. Clearly, taking $(q, \infty)$ as the subcover completes the proof.

$(i)$ connected

In part $(j)$ we prove that $(\Bbb R, \mathcal{T})$ is path-wise connected, hence connected.

$(j)$ path-wise connected

$\Bbb R$ is convex. Given $x, y \in \Bbb R$, the path $f: [0,1] \to \Bbb R$ given by $f(t) = (1-t)x + ty$ is continuous and $f(0) = x$ and $f(1) = y$. Since path-wise connected implies connected we also answered $(i)$.

What is the closure of $\{1\}$ in $(\Bbb R, \mathcal{T})$?

Proof: Since the closure is the smallest closed set containing $\{1\}$, it is clear that it is $(-\infty, 1]$
Any suggestions, corrections, hints and any help, in general, will be tremendously appreciated! Any stylistic improvements in the formatting of the question are also greatly encouraged!
 A: (b) Consider $\bigcup\{(r,\to):r\in\Bbb Q\text{ and }r>\sqrt2\}$.
(f) Consider the set $\mathscr{B}=\{(r,\to):r\in\Bbb Q\}$: is this a base for $\mathscr{T}$?
(g) A small correction: $\{\Bbb R\}$ is a finite cover of the space, as is any finite set of open sets that includes $\Bbb R$. However, $\{(r,\to):r\in\Bbb R\}$ is certainly an open cover with no finite subcover.
(h) Fix $r\in\Bbb R$, and let $\mathscr{U}=\{(q,\to):q>r\}$; $\mathscr{U}$ is an open cover of $(r,\to)$ with no finite subcover. Thus, no member of $\mathscr{T}$ is compact. However, if your definition of neighborhood does not require it to be open, then for any $x\in\Bbb R$ the set $[x-1,\to)$ is a nbhd (though not an open nbhd) of $x$ that is compact: if $x-1\in(r,\to)\in\mathscr{T}$, then $(r,\to)\supseteq[x-1,\to)$, so every open cover of $[x-1,\to)$ has a one-element subcover.
(j) You should say why $f$ is continuous, though I grant that it’s pretty easy.
A: Hint: For $(b)$, use the fact that $\Bbb Q$ is dense in $\Bbb R$. Maybe consider $\sqrt{2}+1/n\to\sqrt{2}$ as $n\to\infty$. For $(g)$, I'm not sure why you say there doesn't exist a finite cover; $\Bbb R$ itself is a perfectly good finite open cover. For $(j)$, I think you need more argument. It seems like you are jumping topologies, which is okay, but you have to argue that the function you've defined is actually continuous via the definition for continuous in topology.
I'm not sure about $(f)$, so I'll leave that for someone else, but hopefully this gets you started on the other parts. I think everything I didn't mention was okay, though.
