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I have this question :

Let $A$ be a non empty set from the space $E$, and define $f_A: E\rightarrow \{0,1\}$ given by $f_A(x)=\begin{cases} 1, ~ x\in A\\ 0 ~ x,\notin A\end{cases}$

where $\{0,1\}$ is given with this topology $\mathcal{P}(\{0,1\})$

How to prove that $f_A$ is continuous at $x_0$ iff $ x_0\notin Fr(A)$

where $Fr(A)=\partial A=\overline{A}\setminus\overset{\circ}{A}$

I know that $f_A ~\text{is continuous at }~x_0 \Longleftrightarrow \forall W\in \mathcal{V}_{f_A(x_0)}, \exists V\in \mathcal {V}_{x_0}, f_{A}(V)\subset W$

If i suppose that $f_A$ is continuous at $x_0\notin A$, then $f_A(x_0)=0$ this means that $\mathcal{V}_{f_A(x_0)}=\{\{0\},\{0,1\}\}$ As $f_A$ is continuous at $x_0$ then there exists $v\in \mathcal{V}_{x}$ such that $f_A(V)\subset \{0\}$

But How to find that $x_0\notin \partial A$ ?

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    Why -1 ????? i begin the proof what is that ??? – 2017-02-15
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    some people downvote for absolutely no real reason. – 2017-02-15

3 Answers 3

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Note that for any $A$ we have that $X = A^\circ \cup \partial A \cup (X \setminus A)^\circ$, where the union is disjoint. The last set also equals $X \setminus \overline{A}$ ,BTW.

For points $p \notin \partial A$ we thus know that $p \in A^\circ$ or $p \in (X \setminus A)^\circ$, and both mean that $f_A$ is locally constant in $p$ (with value $0$ resp. $1$), so continuous at $p$.

If however $p \in \partial A$, every neighbourhood $U$ of $p$ intersects both $A$ and its complement, so $f_A[U] = \{0,1\}$. So the local continuity condition fails for the neighbourhood $\{f(p)\}$ of $f(p)$

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    please if i suppose that $f_A$ is continuous ant $x_0$ and $x_0\notin A$ how to find that $x\in \partial A$ ? – 2017-02-15
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    @Vrouvrou I show $f_A$ continuous at $p$ implies $p \notin \partial A$, this is the contrapositive of the last paragraph! – 2017-02-15
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    i don't understand the proof , you do by contradiction? – 2017-02-15
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    @Vrouvrou it's a direct proof in both directions: $p \notin \partial A$, then $f_A$ continuous at $p$. And if $p \in \partial A$, then $f$ is not continuous at $p$. This proves the equivalence. – 2017-02-15
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    please how you finde that $f_A(V)=\{0,1\}$ – 2017-02-18
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    @Vrouvrou as $U$ contains points of $A$, which map to 1 and points not in $A$ which map to 0. – 2017-02-18
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    please what do you mean by : So the local continuity condition fails for the neighbourhood {f(p)} – 2017-02-18
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    we know that $f_A$ is continuous iff $\forall W\in \mathcal{V}_{f_A(x_0)}, \exists U\in \mathcal{V}_{x_0}, f(U)\subset W$ , you mean that if $x_0\in A$ then $W=\{\{1\}, \{0,1\}\}$ but $f_A(U)=\{0,1\}$, so $f_A(U)\not\subset \{1\}$ ? – 2017-02-18
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    @Vrouvrou almost: if $x_0 \in \partial A$, then for $W = \{f(x)\}$, which is open, for any neighbourhood $U$ of $x_0$, $f_A[U] = \{0,1\} \not \subset W$. So $f_A$ is not continuous at $x_0$, whatever its value. – 2017-02-18
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    @Vrouvrou your last $W = \{\{1\}, \{0,1\}\}$ is not even a neighbourhood of $\{1\}$, as it's not a subset of $\{0,1\}$. It is the set of all neighbourhoods of $1$ in the discrete topology though. – 2017-02-18
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    Yes i must write that $\mathcal{V}_{f_A(x)}=\{\{1\}, \{0,1\}\}$ – 2017-02-18
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    please can you give me a clear answer for this question http://math.stackexchange.com/questions/2162485/necessary-and-sufficient-condition-for-the-continuity-of-a-function – 2017-03-03
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If $x_0\in\partial A$ then for any open $U\subset E$ with $x_0\in U$ we have that $U\cap A\neq\emptyset$, since $x_0\in\overline{A}$, and $U\cap A^c\neq\emptyset$, since $x_0\not\in A^o$. Thus there $f_A(U)=\{0,1\}$. Therefore, if $x_0\in A$, we have that $f_A^{-1}(1)$ is not an open set containing $x_0$. Similarly if $x_0\not\in A$.

Conversely if $x_0\in A^o$ then $f_A^{-1}(1)$ contains an open neighborhood of $x_0$, namely $A^o$, and thus is continuous at $x_0$. You can argue similarly in the case that $x\in (A^c)^o$.

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if $x_0\in Fr(A)$ as $x_0\in \bar{A},\forall V\in\tau_{x_0},V\cap A\neq\emptyset$, so $V\not\subset A^c$. Similarly as $x_0\notin int(A),\overline{\exists V\in\tau_{x_0},V\subset A}\iff\forall V\in\tau_{x_0},V\not\subset A$ and as $f_A(A)=\lbrace 1\rbrace,f_A(A^c)=\lbrace 0\rbrace$ then

$$x_0\in Fr(A)\iff\forall V\in\tau_{x_0},V\not\subset A\land V\not\subset A^c$$$$\iff \not\exists V\in\tau_{x_0},f_A(V)\subset \lbrace 0\rbrace\land\not\exists V\in\tau_{x_0},f_A(V)\subset \lbrace 1\rbrace$$$\iff f_A$ isn't continuous in $x_0$