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Section 32: Problem 2 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Take a point and a point or a closed set contained in (if contains one point only, we are done). For every fix . For every let be the product of with for all . For a subset let . If the product is Hausdorff, and can be separated by basis neighborhoods and . Then and are open (the projection is an open map) and disjoint (if there is a common point then belongs to both and ) neighborhoods of and . Now, if the product is regular, then there are disjoint basis neighborhoods of and , and their projections into are open and disjoint (for the same reason) neighborhoods of and . If the product is normal and is a closed subset of that does not intersect , then and can be separated by disjoint basis neighborhoods whose projections into separate and .
As Yuan says in a comment below, Munkres defines both regular and normal spaces with the additional requirement that singletons are closed. This is actually a common definition now, which guarantees that every normal space is regular etc. So, I missed that point in the original solution. In other words, we need also to show that the space is a space. I leave this as a very easy exercise to the reader.
Of course, I am not leaving you alone. Moreover, here is an alternative way to prove almost all statements in this exercise. We can actually use the hereditary properties of spaces. Indeed, each is homeomorphic to a subspace of the product space (which one?), and so if the product is , Hausdorff or regular, so is . Done. Almost done. Because being a normal space is not a hereditary property, so we still need to show that each is normal if so is the product.