If $Y$ is a Hausdorff topological space, then the value of a continuous function $f\colon X\to Y$ is completely determined by the value of $f$ on a dense subset $E$ of $X$; to see this, suppose $f$ and $g$ are two functions that agree on a dense subset $E$ of $X$, and let $u\in X\setminus E$. If $f(u)\neq g(u)$, then there are open neighborhoods $U$ and $V$ of $f(u)$ and $g(u)$, respectively, such that $U\cap V=\emptyset$. Then $f^{-1}(U)$ is an open neighborhood of $u$, as is $g^{-1}(V)$. Their intersection is an open neighborhood of $u$, and therefore must contain elements of $E$; but then any $e\in E$ in the intersection has $f(e)=g(e)$, with $f(e)\in U$ and $g(e)\in V$, contradicting that $U\cap V=\emptyset$. Therefore, $f(u)=g(u)$, hence $f=g$.
In this situation, it does not matter if $X$ is a complete metric space (or even a metric space); the key is $Y$.
To see that the key is $Y$, consider the extreme case in which $Y$ has the indiscrete topology (the only open sets are the empty set and the set $Y$). Then any function into $Y$ is continuous, so you can make your $X$ anything you want, and have two functions that agree on any subspace you care to specify and yet differ somewhere else.
Added: Note that any metric space is necessarily Hausdorff, so if your maps are between metric spaces, then the property holds (as in the case of Yuval's answer). This because you have the property that $d(x,y)\geq 0$ and $d(x,y)=0$ if and only if $x=y$. Thus, given $x,y\in Y$, $x\neq y$, let $\epsilon=d(x,y)\gt 0$. Then $B(x,\frac{\epsilon}{2})$ and $B(y,\frac{\epsilon}{2})$ are open neighborhoods of $x$ and $y$ (respectively) that are disjoint: if $z$ were in the intersection, then $d(x,z)\lt \frac{\epsilon}{2}$, $d(z,y)\lt\frac{\epsilon}{2}$, and by the triangle inequality we would conclude that $d(x,y)\lt\epsilon$, a contradiction.
Final Addition: See here for a discussion of the converse.
Added 2: A result towards a possible converse: certainly we need some separation on $Y$. Suppose that there exist points $x,y\in Y$, $x\neq y$, such that every open subset that contains $x$ also contains $y$ (so $Y$ could be $T_0$, but cannot be $T_1$). Let $X$ be the Sierpinski space, $X=\{a,b\}$ with topology $\tau = \{\emptyset, \{b\}, X\}$. Let $f,g\colon X\to Y$ be the following maps: $f$ is the constant function that maps everything to $x$; $g$ is the function that maps $a\mapsto x$, $b\mapsto y$. The constantfunction is certainly continuous. For $g$, if $U$ is an open subset that contains $y$ but not $x$, then $g^{-1}(U)=\{b\}$, which is open; so $g$ is continuous. And $g$ and $f$ agree on the dense set $\{a\}$, but are not equal.
I'm still trying to figure out the $T_1$ case (for all $x,y\in Y$, $x\neq y$, there exist open sets $U,V$ such that $x\in U-V$ and $y\in V-U$).
Added 3: Another step: $T_1$ is not sufficient; a colleague came up with this one (I kept trying the cofinite topology on $\mathbb{N}$ and not getting anywhere): take two copies of the real line and identify every point except the origin; this is $Y$. The result is a $T_1$ space, but not Hausdorff since no neighborhoods of the two copies of the origin are disjoint. Now let $X$ be the real line, and let $E = (-\infty,0)\cup(0,\infty)$ be the dense subset. The two obvious injections, one mapping $0$ to the first copy in $Y$ and the other mapping it to the other copy, are both continuous and agree on $E$ but not on all of $X$, so $T_1$ does not suffice for the property. At least, then, in the hierarchy of $T$-spaces, the first level at which we are guaranteed the property is Hausdorff. This does not, however, establish whether the converse property characterises Hausdorff-ness.
Added 4. I almost have the following:
Conjecture. Assuming the Axiom of Choice, the following are equivalent for a topological space $Y$:
- $Y$ is Hausdorff.
- For every topological space $X$, every dense subset $E$ of $X$, and every pair of continuous maps $f,g\colon X\to Y$, if $f$ and $g$ agree on $E$, then $f=g$.
Argument. Suppose that $Y$ is not Hausdorff. If there exist points $x,y\in Y$, $x\neq y$, such that every open neighborhood of $x$ contains $y$, then the map from the Sierpinski space indicated above shows that $Y$ does not have property 2. So we may assume that $Y$ is at least a $T_1$ space. But since $Y$ is not Hausdorff, there exist points $s,t\in Y$, $s\neq t$, such that for every neighborhoods $U$ of $s$ and $V$ of $t$ such that $s\in U-V$ and $t\in V-U$, we have $U\cap V\neq \emptyset$. Let $\mathfrak{U}\_s$ be the family of open neighborhoods of $s$ that do not contain $t$, and let $\mathfrak{V}\_t$ be the family of open neighborhoods of $t$ that do not contain $s$. Let $P=\mathfrak{U}\_s\times \mathfrak{V}\_t$, and partially order $P$ by letting $(U,V)\leq (U',V')$ if and only if $U'\subseteq U$ and $V'\subseteq V$. This makes $P$ into a directed partially ordered set (given any $(U,V),(R,S)\in P$, there exists $(U',V')\in P$ such that $(U,V)\leq (U',V')$ and $(R,S)\leq (U',V')$. Now, for each $(U,V)\in P$, we are assuming that $U\cap V\neq\emptyset$, so using the Axiom of Choice let $y_{(U,V)}\in Y$ be an element of $U\cap V$. Note that $y_{(U,V)}\neq s$ and $y_{(U,V)}\neq t$ for all $(U,V)\in P$.
Now let $X = \{y_{(U,V)}\mid (U,V)\in P\}\cup\{s,t\}$, and give $X$ the induced topology from $Y$, so that the inclusion maps $X\hookrightarrow Y$ is continuous. Note that the set $E=\{y_{(U,V)}\mid (U,V)\in P\}$ is dense in $X$: for every open neighborhood $B$ of $s$ in $X$, there exists an open set $\mathcal{O}\_B\in Y$ such that $\mathcal{O}\_B\cap X = B$; in particular, $\mathcal{O}\_B$ is a neighborhood of $s$; let $\mathcal{V}$ be any open neighborhood of $t$ that does not contain $S$, and let $B'=\mathcal{V}\cap X$; let $\mathcal{U}$ be any open neighborhood of $s$ that does not contain $t$. Then $\mathcal{U}\cap\mathcal{O}\_B$ is open, hence $B'=\mathcal{U}\cap\mathcal{O}\_B\cap X$ is an open subset of $X$ that is contained in $B$. Consider now $y_{(\mathcal{U}\cap\mathcal{O}\_B,\mathcal{V})}$. This is in $\mathcal{U}\cap\mathcal{O}_B\cap\mathcal{V}\cap X \subseteq B'$, and is plainly in $E$. In particular, in $X$ we have that $B'\cap E\neq\emptyset$, and hence $B\cap E\neq \emptyset$. Thus, every open neighborhood of $s$ in $X$ contains points of $E$, hence $x$ lies in the closure of $E$. A symmetric argument holds for $t$. Thus, $E$ is dense in $X$.
Now consider the maps $f,g\colon X\to Y$ defined as follows: $f$ and $g$, restricted to $E$, are the identity; $f(s)=f(t)=s$; and $g(s)=g(t)=t$. I claim that $f$ and $g$ are both continuous. Indeed, let $\mathcal{O}$ be an open set in $Y$. If $\mathcal{O}\cap\{s,t\}=\emptyset$ or $\{s,t\}\subseteq\mathcal{O}$, there is nothing to do: the inverse image under both $f$ and $g$ is just the intersection with $X$, hence open in $X$. So assume without loss of generality that $s\in\mathcal{O}$ but $t\notin \mathcal{O}$. Note that $g^{-1}(\mathcal{O}) = (\mathcal{O}-\{s\})\cap X$, and since $Y$ is $T_1$ removing a single point from an open set results in an open set, so $g^{-1}(\mathcal{O})$ is open. So we just need to show that $f^{-1}(\mathcal{O}) = (\mathcal{O}\cap X)\cup\{t\}$ is open in $X$.
And that is where I am a bit stuck at present. Can anyone either verify or falsify this?
Best Answer
That is there exists a dense set $N$ of $F_1(x)$ such that for each $f\in N$ the set $G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}$ is dense. Note that the set $G_f$ depends on the function $f$ and it may be not the same for distinct $f\in F_1(X)$.
Right.
Note, that we still speak about pairs $(f,\varphi)\in G_n(X)$. And if $x\in E(f)$ and $\varphi(x)=0$ then $(f(x),\varphi(x))=0$, which contradicts $(f,\varphi)\in G_n(X)$.
We are allowed to use $\varphi$ means $(f,\varphi)\in G_n(X)$ for a given fixed $f$. This doesn’t imply that we are allowed to use $\varphi$ with a different function, say $f’\in F_1(X)$. For a particular choice of $f$, when $f(x)=1$ for each $x\in X$, a set $E(f)=\{x\in X :f(x)=0\}$ is empty. Then $G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}=F_{n-1}(X)$. But this does not mean that $G_{f’}=F_{n-1}(X)$ for any function $f’\in F_1(X).$
Remark that in this case also $E(f)=\{x\in X :f(x)=0\}$ is empty, so we cannot "use $\varphi(x)$ with $x\in E(f)$", because there is no such $x$. But if $f’\in F_1(X)$ is an other function then we are allowed to use a function $\varphi\in F_{n-1}$ with the function $f’$, that is $(f,\varphi)\in G_n(X)$ iff $\varphi(x)\ne 0$ for each $x\in E(f’)$.
PS. It seems that the picture is not needed.