Show that the epigraph of $f(x) = 0, x < 0, \text{ and } 1, x\geq 0$ is not closed by examining a plot of the function

Question

Given \begin{equation} f(x) = \begin{cases} 0, x < 0\\ 1, x\geq 0 \end{cases} \end{equation}

With the plot sketched below

Show that the epigraph of this function defined as $$\text{epi}(f(x)) = \{(x,\alpha): f(x) \leq \alpha, x \in \mathbb{R}, \alpha \in \mathbb{R}\}$$ is not closed.

Does anyone know if there is a way to reach this conclusion by just examining the plot?

The example is taken from http://users.ece.utexas.edu/~cmcaram/EE381V_2012F/Lecture_18_Scribe_Notes.final.pdf

Answer 1

The segment $\circ-\bullet$ is clearly a subset of $\operatorname{cl}{\operatorname{epi}f}$, but it is not a subset of $\operatorname{epi}f$.

Added: As an observation, the description of the original image in the notes states

Figure 18.4: Example of a Function Whose Epigraph is Open.

That statement is incorrect, because $\operatorname{epi}f$ is not open either. In fact, as soon as $\operatorname{dom}f\ne\emptyset$, the epigraph of $f$ is guaranteed not to be open: if $f(x)=\alpha\in\Bbb R$, then $(x,\alpha)\in\operatorname{epi}f$ and $\left(x,\alpha-\frac1n\right)\stackrel{n\to\infty}\longrightarrow (x,\alpha)$. But $\left(x,\alpha-\frac1n\right)\notin \operatorname{epi}f$.