0

Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be continuous functions. Suppose that $D ⊆ \mathbb{R}$, and that $D$ is dense in $\mathbb{R}$. Suppose that $f(x) = g(x)$ for every $x ∈ D$. Prove that $f(x) = g(x)$ for every $x ∈ \mathbb{R}$.

Any tips for where I can start here? I'm pulling up blanks for this.

EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately.

ktuggle
  • 33
  • 4

4 Answers4

2

Hint. Let $x \in \mathbb{R}\setminus D$ then, since $D$ is dense in $\mathbb{R}$, there exists a sequence $\{d_n\}_n$ in $D$ such that $d_n\to x$. By your assumption $f(d_n)=g(d_n)$. Now use the continuity of $f$ and $g$ at $x$.

Robert Z
  • 145,942
  • I don't understand the purpose of saying $x\not \in \mathbb{R}\setminus D$, isn't the just a really roundabout way of saying x is in D? – ktuggle Nov 21 '18 at 08:10
  • @ktuggle Sorry, it is $x \in \mathbb{R}\setminus D$, that is $x\not\in D$. – Robert Z Nov 21 '18 at 08:50
1

Hint: Consider $h(x) = f(x)-g(x)$. This should simplify your problem.

Jacky Chong
  • 25,739
0

The set $U:=\{\,f(x)\ne g(x)\mid x\in\Bbb R\}$ is open. We are given that $U\cap D=\emptyset$, but as $D$ is dense, it has non-empty intersection with every non-empty open set. We conclude that $U$ is empty.

Hagen von Eitzen
  • 374,180
0

Hint: Prove the set $ A= \{x \in \mathbb{R} \vert f (x)=g (x)\}$ is closed in $\mathbb {R}$. Clearly, $D \subseteq A $. Taking closure on both sides, we get $\mathbb {R} \subseteq A \implies \mathbb {R}=A $.

cqfd
  • 12,219
  • 6
  • 21
  • 51