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  1. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function that is continuous from the right; that is, for all $a \in \mathbb{R}$,

$\lim_{x \to a^{+}} f(x) = f(a)$

a) Show that $f$ is continuous when viewed as a function from $\mathbb{R}$ with the lower limit topology to $\mathbb{R}$ with the metric topology.

b) Describe the set of functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that are continuous when the domain is given the metric topology and the codomain is given the lower limit topology.

What I am thinking:

Since $f$ is right continuous at $a$:

$\forall \epsilon >0$ $\exists \delta >0$ such that $\forall x$ with $a < x < a +\delta$ then $|f(x) - f(p)| < \epsilon$. So for any open neighborhood of $U$ of $f(a)$, we have an open neighborhood $V$ of $a$ so that $f(V) \subseteq U$. Thus $f: (\mathbb{R},\mathcal{T}_1)\rightarrow (\mathbb{R}, \mathcal{T}_2)$ which goes from the lower limit topology to the metric toplogy.

I'm not sure this answers the question in its entirety and I'm not even sure if I'm correct.

And for part B, I'm just genuinely confused, I think I have an idea but I'm pretty sure I'm wrong, so any help is really appreciated.

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Your solution for [a)] is correct. Hint for [b)]:

A continuous image of a connected set is connected, what are the connected sets of $\mathbb{R}_l$ (the reals with the lower limit topology)?

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  • $\begingroup$ uhh only one point sets? $\endgroup$ – user130306 Mar 18 '18 at 3:35
  • $\begingroup$ @user130306 yes, exactly $\endgroup$ – j3M Mar 18 '18 at 7:01

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