# prove that a piecewise function at an isolated point is continuous

Prove that $$f$$ is continuous at $$x = 3$$

My proof

We perform a few small calculations to determine that the function is: $$f(x):=\left\{ \begin{array}{cc} -x+2 & \text{ if } x\leq 2\\ 2 & \text{ if } x=3\\ 1 & \text{ if } x \geq 4 \end{array} \right.$$ The domain of the function is $$(-\infty, 2] \cup \{ 3 \} \cup [4, \infty)$$. \

Before demonstrating, let's recall the definition of continuity:

$$(X, d_x),(Y, d_y),$$ Metric spaces, $$f: E\subseteq X \to Y$$, $$x_0 \in D(f)$$. We say that $$f$$ is continuous at $$x=x_0$$ if and only if $$\forall \epsilon > 0, \exists \delta >0,$$ such that: if $$x \in D(f)$$ and $$d_x( x, x_0) < \delta,$$ then $$d_y(f(x), f(x_0)) < \epsilon$$.

and the definition of isolated point:

Given a set $$A$$ and a point $$x \in A$$, $$x$$ is an isolated point of $$A$$ if and only if $$x \in A$$ and $$\exists r >0$$ such that $$B_r( x) \cap A=\{ x \}$$ \

Continuing with the demonstration, we note that the point $$x=3$$ is an isolated point, for this reason we cannot approach it from the right or from the left. Then let $$x_0 \in D(f)$$, more specifically $$x_0 = 3$$. Let $$\delta > 0$$ and $$x \in D(f)$$ satisfy $$d(x, x_0) < \delta$$. By definition of distance we have $$(x_0-\delta, x_0+\delta)$$ and since $$x_0$$ is an isolated point we can write it: $$(x_0-\delta, x_0+\delta) \cap D(f) =\{ x_0 \}$$ Now let's choose any $$\epsilon>0$$. Then we have that for any $$x \in (x_0-\delta, x_0+\delta) \cap D(f) =\{ x_0 \}$$ : $$|f(x)-f(x_0)|=|f(x_0)-f(x_0)|= |2-2|= 0 < \epsilon .$$ This satisfies the continuity condition at $$x_0$$, and the limit in this case is trivially equal to $$f(x)$$.

is this test correct?

• Weird exercise you were given there, but your proof is correct. May 25, 2023 at 14:33
• Probably an exercise to show that odd things can be true when the domain is disconnected. What would have been a jump discontinuity isn't so because the domain itself is disconnected. May 25, 2023 at 14:36
• @0XLR Thanks for your answer, yes I think he put it for that, yes it's weird like all the homework my profesor gives me May 25, 2023 at 18:17
• @BrunoB Thank you for your answer, I really need to get a good grade on this assignment May 25, 2023 at 18:18
• If the domain of a function is discrete, the function is always continuous. That means, for example, any sequence is continuous. See here. May 25, 2023 at 18:24