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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?

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  • 1
    $\begingroup$ Weird exercise you were given there, but your proof is correct. $\endgroup$
    – Bruno B
    May 25, 2023 at 14:33
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    $\begingroup$ 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. $\endgroup$
    – balddraz
    May 25, 2023 at 14:36
  • $\begingroup$ @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 $\endgroup$
    – larous25
    May 25, 2023 at 18:17
  • $\begingroup$ @BrunoB Thank you for your answer, I really need to get a good grade on this assignment $\endgroup$
    – larous25
    May 25, 2023 at 18:18
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    $\begingroup$ If the domain of a function is discrete, the function is always continuous. That means, for example, any sequence is continuous. See here. $\endgroup$
    – durianice
    May 25, 2023 at 18:24

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