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I think my definition of continuous must be wrong, or I am doing something wrong in answering this question. Can someone tell me what is wrong in my proof?

where $F$ is the characteristic function, and $\Delta$ is the cantor set, Show that $F_\Delta:R\rightarrow R$ is discontinuous at each point of $\Delta$

The characteristic function is $0$ if $x\notin \Delta$ and 1 if $x \in \Delta$,

Definition of continuous from text :

$x,a \in A$, where $A$ and $B$ are some sets, $f:A\rightarrow B$

$\forall \epsilon > 0 \,\exists \,\delta >0$ S.T. $|f(x)-f(a)|<\epsilon$ whenever $|x-a|<\delta$

If you consider the Cantor function as the intersection of intervals, $I_n$, where each $I_n$ is broken into a union of intervals $K_{i}$, then $\forall \,X \in \Delta$, $x\in I_nK_i$. Basically for any $n$, $x$ is in some $I_nK_i$. Each of these $I_nK_i$ is of length $\frac{1}{3^n}$. If you choose $x_n$ to be the right endpoint of each of these $I_nk_i$, then $x=lim_{n\rightarrow \infty}\{x_n\}$

Since each point in $\Delta$ is the limit of a sequence {$x_n\}_0^\infty$, where each $x_n$ is in $\Delta$, then $\forall \, \delta \, \exists \,n$ S.T. $|x_n-x|<\delta$.

Thus, $\forall \epsilon > 0,\, |f(x_n)-f(x)|=1-1=0<\epsilon$

So $\Delta$ should be continuous by the definition above.

But the questions answer states $\Delta$ is not continuous because the Cantor function contains no open intervals the characteristic function is open.

What am I missing, is my definition of continuous wrong?

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  • $\begingroup$ You are using the sequential characterization of continuous, as I read it. Under this definition, $f$ is continuous if for ANY sequence $x_n$ converging to $x$, $f(x_n)\rightarrow f(x)$. You have chosen a very particular sequence that happens to satisfy $f(x_n)\rightarrow f(x)$ but haven't proven continuity. $\endgroup$ – user25959 Dec 8 '18 at 18:27
  • $\begingroup$ And btw it would certainly be easier to use the fact that $\Delta$ contains no open intervals to show discontinuity of $f$. This means that any $x \in \Delta$, whereas $f(x)=1$, you can find values of $0$ no matter how close you get to $x$. $\endgroup$ – user25959 Dec 8 '18 at 18:29
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One characterization of continuity of a function $f$ at a point $x$ is that for all sequences $(x_n)$ such that $x_n\to x$ we have that $f(x_n)\to f(x)$. You have only shown this implication for one choice of $(x_n)$. Your proof shows that any point in the cantor set is the limit of points in the cantor set. But it is also the case every point in the cantor set is the limit of points not in the cantor set.

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