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Let $(X,{\mathcal T}_X)$ and $(Y,{\mathcal T}_Y)$ be topologiclal spaces, and let $f,g:(X,{\mathcal T}_X)\to(Y,{\mathcal T}_Y)$ be continuous maps.

Define the equality set as $$E(f,g) = \{x\in X \ | \ f(x) = g(x) \}$$

I have worked out that if $(Y,{\mathcal T}_Y)$ is Hausdorff, then $E(f,g)$ is ${\mathcal T}_X$-closed (see this answer).

In order to get a better understanding I am trying to find examples of continuous maps $f,g$ with $E(f,g)$ not closed. My understanding is that this only occurs for some maps where the target space $(Y,{\mathcal T}_Y)$ is not Hausdorff.

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elasticSkis
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2 Answers2

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Define the Sierpinski space S to be the set $X = \{0,1\}$, together with the topology $\mathscr{T}_X = \{\emptyset, \{1\}, \{0,1\}\}$. There clearly $S$ is not Hausdorff, as we cannot isolate $0$ and $1$ using two disjoint open sets. Let $\mathbb{R^*_+}$ be equipped with its natural topology. Define $f,g : \mathbb{R^*_+} \rightarrow S$ such that:

(1) $f(x) = 1, \forall x \in \mathbb{R^*_+}$

(2) $g(x) = 1$ if $x \in ]0,1[$ and $g(x) = 0$ if $x \geq 1$.

$f$ is constant therefore continuous, and g is continuous as well. If fact, it amounts to prove that the inverse image of any open set of $S$ is open in $\mathbb{R^*_+}$. Now, we have: $g^{-1}(\{1\}) = ]0,1[$ (which is open in $\mathbb{R^*_+})$, $g^{-1}(S) = \mathbb{R^*_+}$ (open) and $g^{-1}(\emptyset) = \emptyset$ (open as well). Therefore $g$ is indeed continuous, and we have:

$E(f,g) = ]0,1[$, which is not closed in $\mathbb{R^*_+}$, therefore providing us a counterexample.

Hermès
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An example where $Y$ is a $T_1$ space but not Hausdorff . Let $T_R$ be the usual topology on the reals $R$. Let $Q$ be the rationals.Let $Y=Q\cup ((R$ \ $Q)\times \{0,1\}).$

For $b\subset R$ let $b^*=(b\cap Q)\cup ((b$ \ $Q)\times \{0,1\}).$ Let $B=\{b^*$ \ $c : b\in T_R$ and $c$ is finite $\}.$ Then $B$ is a base for a topology $T_Y$ on $Y.$ And $(Y,T_Y)$ is a $T_1$ space but not a $T_2$ space.

Define $f:R\to Y$ and $g:R\to Y$ by :

(i). $f(q)=g(q)=q$ if $q\in Q .$

(ii) If $r\in R$ \ $Q$ then $f(r)=(r,0)$ and $g(r)=(r,1).$

Then $f$ and $g$ are continuous, as can be verified by checking that $f^{-1}d$ and $g^{-1}d$ are open in $R$ for each $d\in B.$ But $\{x: f(x)=g(x)\}=Q$ which is not closed in $R.$

Another example (in which $Y$ is not $T_1$): Let $T_Y$ be the coarse (anti-discreet) topology on $Y, $ where $Y$ has at least 2 members. Any function into $Y$ is continuous. Let $A$ be any non-closed subset of a space $X$ . Let $y_1\in Y. $ Let $f(x)=y_1$ for all $x\in X.$ Let $g(x)=y_1$ for $x\in A$ and $g(x)\in Y$ \ $\{y_1\}$ for $x\in X$ \ $A.$ Then $\{x:f(x)=g(x)\}=A.$

DanielWainfleet
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