# Product of a compact set and a singleton is compact proof

It's before we prove that 'Product of two compact sets is compact'. Let $$S$$ be an open cover of $$X \times \{\bullet\}$$ where $$X$$ is compact. Then $$\pi_1(S)$$ is an open cover of $$X$$ so there is a finite subcover then pick one $$A_n$$ from $$S$$ corresponding to each $$\pi_1(A_n)$$ but I think it may not cover $$X \times \{\bullet\}$$. How to complete the proof?

Edit There are topological spaces $$X$$(which is compact), $$Y$$ and the product topology on $$X \times Y$$ is given by the subbase $$U \times V$$ where $$U$$ is open in $$X$$ and $$V$$ is open in $$Y$$. Pick an element $$\bullet \in Y$$. Let $$S$$ be an open cover of $$X \times \{\bullet\}$$. Then $$\pi_1(S)$$ is an open cover of $$X$$ so there is a finite subcover then pick one $$A_n$$ from $$S$$ corresponding to each $$\pi_1(A_n)$$ but I think it may not cover $$X \times \{\bullet\}$$. For example let $$Y$$ be a $$T_1$$ space and pick another element $$\bullet\bullet \in Y$$. There exists an open set $$W$$ containing $$\bullet\bullet$$ but not $$\bullet$$. Lets construct an open cover of $$X \times \{\bullet\}$$ $$S \cup \{B \times W \mid B \in \pi_1(S)\}$$ Now we can apply this open cover to upper proof, and when we are picking $$A_n$$ from $$S$$ corresponding to each $$\pi_1(A_n)$$, we may pick all the sets from $$\{B \times W \mid B \in \pi_1(S)\}$$ so in fact it doesn't cover $$X \times \{\bullet\}$$. Am I misunderstanding something?

It will cover $$X \times \{∙\}$$.

You have a finite collection $$\pi_1(A_1),\dots,\pi_1(A_n)$$ of open subsets of $$X$$ that cover $$X$$. Choose any $$(x,∙) \in A_i$$. Then $$x \in X$$, so there exists $$A_i$$ such that $$x \in \pi_1(A_i)$$. This means there exists $$(y,z) \in A_i\subseteq X \times \{∙\}$$ such that $$\pi_1(y,z)=x$$. But the only $$z \in \{∙\}$$ is $$∙$$ itself, and $$\pi_1(y,z)=y$$ by definition. So we have $$(x,∙) \in A_i$$.

However, you should also justify why $$\pi_1(A_i)$$ is open. In general the image of an open set is not necessarily open, but it will hold for projections.

• I edited the question sorry for short explanations. – Ris Jan 26 at 14:26
• In your edit, is $Y$ compact? – kccu Jan 26 at 14:32
• Not necessarily. – Ris Jan 26 at 14:34
• Then $X \times Y$ is not necessarily compact. (Also, you are using $X$ to denote an element of $\pi_1(S)$, but $X$ is also the name of your compact space...) – kccu Jan 26 at 14:34
• Oh I edited it to $B$. Then $X \times \{ \bullet \}$ is also not necessarily compact? – Ris Jan 26 at 14:36

$$\pi_1$$ is a homeomorphism between X×{•} and X.