Prove $f(x) = g(x)$ for all $x \in \overline{A}$.

Question

Let $f,g : X \rightarrow Y$ be continuous functions from a topological space $X$ into a Hausdorff space $Y$. Let $A \subset X$ and suppose $f(x)=g(x)$ for all $x \in A$. Prove $f(x) = g(x)$ for all $x \in \overline{A}$.

$\textbf{Definition:}$ $X$ is $\textbf{Hausdorff}$ means for each distinct pair of $x_1,x_2 \in X$, $\exists \, \text{disjoint open} \, U,V \subset X$ with $x_1 \in U$ and $x_2 \in V$ such that $U \cup V = X$.

$\textbf{Definition:}$ A function $f: X \rightarrow Y$ where $X,Y$ are topological spaces is $\textbf{continuous}$ means $\forall \, \text{open} \, V \subset Y$, $f^{-1}(V)$ is open in $X$.

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I am unsure how to start this problem off. I wrote the two given definitions to try and get a clearer view on the goal. Any hint on how to begin this would be appreciated.

Answer 1

Take $x\in\overline A$ and suppose that $f(x)\ne g(x)$. Since $Y$ is Hausdorff, there are neighborhoods $V_f$ and $V_g$ of $f(x)$ and of $g(x)$ respectively such that $V_f\cap V_g=\emptyset$. Since $f$ is continuous, $f^{-1}(V_f)$ and $g^{-1}(V_g)$ are neighborhoods of $x$. Therefore, $f^{-1}(V_f)\cap g^{-1}(V_g)$ is also a neighborhood of $x$ and so it contains some $a\in A$. But then $f(a)\in V_f$ and $g(a)\in V_g$, which is impossible, since $f(a)=g(a)$ and $V_f\cap V_g=\emptyset$.

Answer 2

Hint : What do you know about the diagonal of a an hausdorff space ?

Answer 3

I would use the limit point definition of continuity: $f$ is continuous iff $f$ preserves limits of all nets.

Answer 4

Take a point $x$ such that $f(x)\neq g(x)$. Then because $Y$ is Hausdorff, there are open disjoint subsets $U\ni f(x)$ and $V\ni g(x)$ of $Y$.

Take the intersection $f^{-1}(U)\cap g^{-1}(V)$. It is open. It contains $x$. There is no point in this intersection where $f$ and $g$ agree. Is it possible for $x$ to lie in $\overline A$?

Answer 5

Let $y_1=f(x)$ and $y_2=g(x)$ where $x\in\overline A$.

If $U_1\ni y_1$ and $U_2\ni y_2$ are any open neighbourhoods and $f,g$ are continuous, the set $V=f^{-1}(U_1)\cap g^{-1}(U_2)$ is an open neighbourhood of $x$. As $x\in \overline A$, $V$ must intersect $A$. So let $a\in V\cap A$. Then $U_1\ni f(a)=g(a)\in U_2$, i.e., $U_1$ and $U_2$ are not disjoint.

Wec conclude that $y_1$ and $y_2$ cannot be separated in $Y$. As $Y$ is Hausdorff, this means that $y_1=y_2$.