Prob. 5, Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The set of points of equality of two continuous mappings into a Hausdorff space is closed

Here is Prob. 5, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Let $$f, g \colon X \rightarrow Y$$ be continuous; assume that $$Y$$ is Hausdorff. Show that $$\big\{ x \vert f(x) = g(x) \big\}$$ is closed in $$X$$.

My Attempt:

Let $$X$$ and $$Y$$ be any topological spaces, and let $$f \colon X \rightarrow Y$$ and $$g \colon X \rightarrow Y$$ be continuous mappings; suppose that $$Y$$ is a Hausdorff space. Let us put $$S \colon= \big\{ \, x \in X \, \vert \, f(x) = g(x) \, \big\}. \tag{Definition 0}$$ We need to show that this set $$S$$ is closed in $$X$$. For this we show that the set $$X \setminus S$$ is open in $$X$$.

Let $$p$$ be any point of $$X \setminus S$$. Then $$p \in X$$ and $$f(p) \neq g(p)$$, that is, $$f(p)$$ and $$g(p)$$ are two distinct points of the Hausdorff space $$Y$$, which implies that there exist disjoint open sets $$V_f$$ and $$V_g$$ of $$Y$$ containing $$f(p)$$ and $$g(p)$$, respectively. Let us now put $$U_f \colon= f^{-1} \left( V_f \right) \qquad \mbox{ and } \qquad U_g \colon= g^{-1} \left( V_g \right). \tag{Definition 1}$$ Then as the maps $$f \colon X \rightarrow Y$$ and $$g \colon X \rightarrow Y$$ are continuous, as $$V_f$$ and $$V_g$$ are open sets in $$Y$$ containing $$f(p)$$ and $$g(p)$$, respectively, so both the sets $$U_f$$ and $$U_g$$ are open sets in $$X$$ containing the point $$p$$. Let us now put $$U_p \colon= U_f \cap U_g. \tag{Definition 2}$$ Then $$U_p$$ is an open set of $$X$$ containing the point $$p$$.

Moreover, if $$x \in U_p$$, then we have $$x \in U_f$$ and $$x \in U_g$$, that is, $$x \in X$$ for which $$f(x) \in V_f$$ and $$g(x) \in V_g$$, and as $$V_f$$ and $$V_g$$ are disjoint, so we can conclude that $$f(x) \neq g(x)$$, and thus $$x \in X \setminus S$$. Thus it follows that $$U_p \subset \, X \setminus S.$$

Thus we have shown that for any point $$p \in X \setminus S$$, there exists an open set $$U_p$$ of $$X$$ such that $$p \in U_p$$ and $$U_p \subset X \setminus S$$. Thus $$X \setminus S$$ is an open set in $$X$$, by Prob. 1, Sec. 13, in Munkres. Hence $$S$$ is a closed set in $$X$$.

Is this proof correct? If so, is it clear enough for any novice student of topology? Or, is it incorrect somewhere or unclear?

• Seems more than clear to me. – Sahiba Arora May 18 '20 at 20:32
• Step by step, very clear. – Peter Szilas May 18 '20 at 21:07

1 Answer

Quite clear, I agree with the comments on that point.

As an additional proof idea:

I think Munkres also has an exercise (or maybe a theorem) where he shows that $$Y$$ is Hausdorff iff $$\Delta_Y = \{(y,y): y \in Y\}$$ is closed in $$Y \times Y$$ in the product topology.

And if $$f,g: X \to Y$$ are continuous, so is $$f \nabla g: X \to Y \times Y$$ defined by $$(f \nabla g)(x)=(f(x), g(x))$$ e.g. because $$\pi_1 \circ (f \nabla g) = f$$ and $$\pi_2 \circ (f \nabla g) = g$$ and the universal property of continuity of product maps.

Then note that by definition $$S=(f \nabla g)^{-1}[\Delta_Y]$$

and is thus closed in $$X$$ for a Hausdorff space $$Y$$.