Munkres Topology Solutions Chapter 5 Info

Proof sketch. Let $X_\alpha$ compact. Let $(x_i) i\in I$ be a net in $\prod X \alpha$. For each $\alpha$, the projection $\pi_\alpha(x_i)$ is a net in $X_\alpha$; it has a convergent subnet. Use a diagonal argument (or Zorn’s lemma on index sets) to extract a subnet converging coordinatewise. In product topology, coordinatewise convergence = convergence. Thus product is compact. □

While the first part of the chapter focuses on the product, the latter sections (often grouped in Chapter 5 or spilling into Chapter 6 depending on the edition structure) deal with the interplay between and Separation Axioms ($T_1, T_2, T_3, T_4$). munkres topology solutions chapter 5

If you are searching for solutions, you have likely already read the chapter. But let’s refresh the essentials. Proof sketch

Let $X = \mathbbR$ (not compact) and $Y = \mathbbR$. Let $N = < 1 $. The slice $0 \times \mathbbR$ is contained in $N$ (since $0\cdot y = 0$). But no neighborhood $W$ of $0$ in $\mathbbR$ satisfies $W \times \mathbbR \subset N$, because for any $W$ containing $0$, pick a large $y$ and $x \in W$ with $x$ small but nonzero, then $|xy|$ can exceed 1. So tube lemma fails. For each $\alpha$, the projection $\pi_\alpha(x_i)$ is a

Let $X$ be a Tychonoff space. Show that if $f: X \to \mathbbR$ is bounded and continuous, then $f$ extends to $\beta X$.