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Let $f:\mathbb R\to \mathbb R$ be any function, with $G$ denoting the graph of $f$. My task is to show the following:

There exists a countable dense subset $D$ of $\mathbb R$ such that for each $t\in \mathbb R$, there exists an increasing sequence of $t_n\in D$ such that $t_n\to t$ and $f(t_n)\to f(t)$.

My idea is as follows: the graph $G$ is a subset of $\mathbb R^2$ which is separable, and hence $G$ is separable. Take $E$ to be a countable dense subset of $G$ and we are almost done. However, how can I take such sequence $t_n$ to be increasing?

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    I suspect that just having any old countable dense subset $E \subset G$ you are not "almost done". Mostly likely you will need to choose $E$ more carefully amongst the many countable dense subsets of $G$ before you find the one that will resolve your task. – 2017-02-16
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    Thank you for your comment. By saying "almost done" I mean if such $E$ is chosen, then for any $(t,f(t))\in G$, there exists $(t_n,f(t_n))\in E\subseteq G$ such that $(t_n,f(t_n))\to (t,f(t))$. Take $D$ to be the projection of $E$ to $\mathbb R$ and $D$ is countable and dense in $\mathbb R$ .Where was I wrong? – 2017-02-16
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    What I'm saying is that you may be trying to prove something false. The theorem says **there exists** a countable dense set $D \subset \mathbb{R}$ with those properties. It does not say **every** countable dense set $D \subset \mathbb{R}$ has those properties. Careful choice of the correct $D$ is critical to the proof. Choosing the wrong $D$ will sabotage the proof. – 2017-02-16

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