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Background:

Suppose that we have a function which is a real-valued piecewise continuous algebraic function $f : \mathbb{R}^{n} \longrightarrow \mathbb{R}$. By that we mean that: $$f(x) = \begin{cases} f_{1}(x) \mbox{ if }~ x \in S_{1} \\ f_{2}(x) \mbox{ if }~ x \in S_{2} \\ f_{3}(x) \mbox{ if }~ x \in S_{k} \end{cases} $$

Where $x = (x_1,...,x_n)$. Here:

  1. Each $f_{i}$ is an real-value algebraic function, meaning there exits a polynomial $p$ such that $p(x,f(x)) = 0$
  2. Each $f_{i}$ is continuous on $S_{i}$, meaning that we only really need to worry about continuity in boundaries of the $S_{i}'s$
  3. $S_{1}\cup S_{2} \cup S_{3}$ is connected.

These arise when considering semi-algebraic functions.

Question: If $f$ has a connected graph, does that imply $f$ is continuous.

I understand that having a connected graph is not enough to imply continuity in general, as the topologist sine curve shows. I'm just hoping that the lack of oscillatory behavior in a function which is piecewise algebraic will help. Any pointers would be welcome.

Thanks in advance.

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1 Answer

up vote 1 down vote accepted

No, your result is false, and here is a couterexample:

consider $S_1=(-\infty,0]\times{\mathbb{R}}$, $S_2=(0,\infty)\times{(-\infty,0]}$, and $S_3=(0,\infty)\times{(0,\infty]}$, define $f:\mathbb{R}^2\rightarrow{\mathbb{R}}$ by

  • $f(x,y)=x$ if $(x,y)\in{S_2}$
  • $f(x,y)=0$, otherwise

Then the graph of f is connected; do the draw, but f is not continuous, for $f(0,-1)=0$ and $f(\epsilon,-1)=\epsilon$ for all $0<\epsilon<1$.But the restrictions of $f$ to $S_1,S_2,S_3$ are algebraic and $S_1\cup{S_2\cup{S_3}}$ is connected. Intuitively I think the result is true for $n=1$.

Greets

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But isn't $(\epsilon,0)\in S_2$ (for $0<\epsilon<1$), so $f(\epsilon,0)=\epsilon$? It is true that $f(1,\epsilon)=0$ for these $\epsilon$ though, so I think your example still works. –  Matt Pressland Oct 24 '12 at 9:32
    
Thank you very much. I think Matt Pressland comments is correct. –  acyrl Oct 24 '12 at 17:49
    
thanks, I just corrected the justification that $f$ is not continuous, but it is kind of obvious though –  Camilo Arosemena Oct 24 '12 at 22:25
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