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Can anybody help please help me, I have to answer this problem in topology:

"Let $f$ and $g$ be continuous functions from the topological space $T$ into $\mathbb{R}$, with the usual topology. Show that $\{ x \in T \:| \:f(x) = g(x) \}$ is closed.

Also show that if $f(x)=g(x)$ for all x in some dense subset of $T$, then $f=g$ for all $x$ in $T$."

I hope someone can help me with this! Thank you.

Dan Rust
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mathgirl310
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2 Answers2

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Let $h:T\to\Bbb R$ be defined by $h(x)=f(x)-g(x)$. You should be able to show that $h$ is continuous. Your set is precisely the preimage under $h$ of the closed set $\{0\}$, so is closed by continuity of $h$.

If $S$ is a dense subset of $T$, then the closure of $S$ is $T$. But the closure of $S$ is the smallest closed subset of $T$ that contains $S$, and so if $S$ is a subset of the closed set in the first part (that is, if $f,g$ agree on $S$), then $T$ is the set from the first part, meaning $f=g$.

Cameron Buie
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  • Ah, silly me. My suggestion above was unnecessary, but would work for the general case where $f,g\colon T\to Y$, $Y$ Hausdorff. – Ted Shifrin Jun 01 '13 at 20:01
  • @Ted: Indeed, your method is more broadly applicable, but here, it felt a bit like swatting a fly with a Buick, to me. ;-) – Cameron Buie Jun 01 '13 at 20:03
  • I agree. I still would rather not give complete solutions when it seems 100% likely it's homework :P – Ted Shifrin Jun 01 '13 at 20:04
  • Thank you so much, this is really helpful :-) But what about the part where I have to show that it it a closed set? I know I probably have to show that the complement is open, but how do I do that? – mathgirl310 Jun 01 '13 at 21:22
  • @mathgirl310: Which "it" are you talking about showing to be closed? ${0}$? The preimage of ${0}$ under $h$? The closure of $S$? – Cameron Buie Jun 01 '13 at 21:24
  • I am talking about the set {x∈T|f(x)=g(x)} :) – mathgirl310 Jun 01 '13 at 21:47
  • @mathgirl310: Gotcha. Given an arbitrary topological space $T$, what is your definition of continuity for functions $T\to\Bbb R$? – Cameron Buie Jun 01 '13 at 21:49
  • I suspect it is one of the following: (1) For all $x\in T$ and all $\epsilon>0$, there is some open $U\subseteq T$ with $x\in U$ such that $|h(x)-h(y)|<\epsilon$ whenever $y\in U$. (2) For any open $x\in T$ and any open $V\subseteq\Bbb R$ such that $h(x)\in V$, there is some open $U\subseteq T$ with $x\in U$ such that $h(U)\subseteq V$. (3) The pre-image under $h$ of any $\Bbb R$-open set is $T$-open. (4) The pre-image under $h$ of any $\Bbb R$-closed set is $T$-closed. Definition (4) is the one I hinted at in my answer. You should be able to show they are all equivalent. – Cameron Buie Jun 01 '13 at 22:10
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A nicer result can be seen as following:

Theorem: For any pair $f,g$ of continuous mappings of a space $X$ into a Hausdorff space $Y$, the set $\{x\in X: f(x)=g(x)\}$ is closed in $X$.

Proof: We shall show that the set $A=\{x \in X: f(x)\not=g(x)\}$ is open. For every $x \in A$ we have $f(x)\not=g(x)$; hence there exist in $Y$ open sets $U_1, U_2$ such that $f(x)\in U_1$, $g(x)\in U_2$ and $U_1\cap U_2=\emptyset$. The set $f^{-1}(U_1) \cap g^{-1}(U_2)$ is a neighbourhood of $x$ contained in $A$.

And hence,

Corollary: For any pair $f,g$ of continuous mappings of a space $X$ into a Hausdorff space $Y$ with $f(x)=g(x)$ for any $x\in S$, where $S$ a dense subset of $X$, then for any $x\in X$, $f(x)=g(x)$.

Proof: It is clearly to see that $S \subset \{x\in X: f(x)=g(x)\}$. By the above theorem, we can conclude that $\overline{S} \subset \{x\in X: f(x)=g(x)\}$, and hence $X = \{x\in X: f(x)=g(x)\}$.

Paul
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