5
$\begingroup$

Let $(X,d)$ and $(Y,d')$ be metric spaces, and let $D$ be a dense subset of $X$. Show that:

If $f:X\to Y$ and $g:X\to Y$ be continuous, then the set $\{x\in X\mid f(x)=g(x)\}$ is closed.

$\endgroup$
1
  • 1
    $\begingroup$ How the dense $D$ does come in the picture? $\endgroup$ – Berci Apr 8 '13 at 17:14
2
$\begingroup$

Where does the set $D$ come into picture? This is an irrelevant piece of information needed to prove the question. Below is a hint.

HINT

Using the fact that continuity implies sequential continuity, try to prove that the set $\{x \in X \vert f(x) = g(x)\}$ contain its limit points.

$\endgroup$
2
$\begingroup$

In a metric space $X$ we have

  1. $S\subseteq X$ is closed iff it is closed under limit of sequences, i.e., for all sequences $(s_n)\subseteq S$, $\ \exists\lim(s_n)\Rightarrow \lim(s_n)\in S$.
  2. $f:X\to Y$ is continuous iff it is sequentially continuous (i.e., preserves limits of sequences).

Now, let $S:=\{x\in X\,\mid\,f(x)=g(x)\}$, and let $(s_n)\subseteq S$ be a convergent sequence, say $s_n\to x\in X$. Then, by continuity, we have $$f(x)=\lim f(s_n)=\lim g(s_n)=g(x)\,.$$ So, $x\in S$.

$\endgroup$
2
$\begingroup$

One of the axioms of a metric space is $d(a,b)=0$ if, and only if, $a=b$.

Define a map $\phi : X \to \mathbb{R}$ as follows: $\phi : x \mapsto d'(f(x),g(x))$. Clearly, $\phi(x)=0$ if, and only if, $f(x)=g(x)$. The aim is to show that $\phi^{-1}(0)$ is closed in $X$. By assumption, $f$ and $g$ are continuous. As with any metric space $d' : Y \times Y \to \mathbb{R}$ is (uniformaly) continuous. Hence $\phi$ is the composite of two continuous maps.

\begin{array}{ccccc} X &\stackrel{f\times g}{\longrightarrow}& Y \times Y &\stackrel{d'}{\longrightarrow}& \mathbb{R} \\ \\ x &\longmapsto& (f(x),g(x)) &\longmapsto& d'(f(x),g(x)) \end{array}

Since $\{0\}$ is closed in $\mathbb{R}$ it follows that $\phi^{-1}(0) \subseteq X$ is closed in $X$.

$\endgroup$
0
$\begingroup$

A Hausdorff space (also called a $T_2$ space). is a topological space $S$ in which any two distinct points $x,y\in S$ have disjoint neighborhoods: There are open sets $U,V$ with $x\in U,\;y\in V$ and $U\cap V=\phi.$

THEOREM: If $Y$ is Hausdorff and $f:X\to y,\;g:X\to Y$ are continuous then $A=\{p\in X:f(p)=g(p)\}$ is closed in $X.$

PROOF:Suppose not. Let $p\in (Cl_X A)\backslash A.$ So $f(p)\ne g(p).$ Let $U,V$ be disjoint open sets in $Y$ with $f(p)\in U$ and $g(p)\in V.$

Since $f$ and $g$ are continuous there are open sets $U^*,\; V^*$ in $X$, with $p\in U^*,\; p\in V^*$, satisfying $f(U^*)\subset U$ and $g(V^*)\subset V.$

Now for any $q\in U^*\cap V^*$ we have $f(q)\ne g(q)$ because $f(q)\in U,\; g(q)\in V$ and $U\cap V=\phi.\;$

So $ U^*\cap V^*$ is disjoint from $A,$ and is open in $X,$ and contains $p.$ But this implies $p\not \in Cl_X A,$ a contradiction.QED.

Observe that a metric space $(Y,d')$ is Hausdorff ,for if $x,y$ are distinct points of $Y,$ let $U=B_{d'}(x,r)$ and $V=B_{d'}(y,r)$ where $r=d'(x,y)/2.$

COROLLARY: If $Y$ is Hausdorff and $f:X\to Y,\; g:X\to Y$ are continuous, and if $A=\{p:f(p)=g(p)\}$ is dense in $X,$ then $f=g.$ Becuase, by the theorem we have $ Cl_X A=A$ and by the hypothesis of denseness we also have $Cl_X A=X.$

$\endgroup$
1
  • $\begingroup$ Please use the 'enter' key more; your answer as it is written is far too dense for me to want to read it. $\endgroup$ – Eric Stucky Feb 7 '16 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.