Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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**Pseudocompactness**

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space is a pseudocompact space if every continuous real-valued function defined on is bounded, i.e., if is a continuous function, then is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space where is a maximal almost disjoint family of subsets of (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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**The exercise**

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

*Theorem 1*

Let be a compact space. Let be a one-to-one continuous function from onto a space . Then is a homeomorphism.

*Theorem 2*

Let be a pseudocompact space. Let be a one-to-one continuous function from onto where is a separable and metrizable space. Then is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map as described in Theorem 2, the domain must be compact. Then Theorem 1 will finish the job.

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**Reference**

- Arkhangelskii A. V.,
*Topological Function Spaces*, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992. - Arkhangelskii A. V., Ponomarev V. I.,
*Fundamental of general topology: problems and exercises*, Reidel, 1984. (Translated from the Russian).

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Hi Dan Ma â

Thanks for including me on the mailing list of your blog.

Iâd not heard of pseudo-compactness, but had made up an exercise for real analysis students (undergrads) thatâs sort of similar. If every continuous real valued function on a metric space is bounded then the space is compact. The proof is: non-compact implies thereâs a sequence that accumulates nowhere. Put a closed neighborhood around each point in the sequence so the neighborhoods are disjoint. Use the metric to define a function with support on the nth neighborhood that has value n at the middle. Take the sum of these functions. Itâs continuous and unbounded.

Similar exercises with the same proof: the metric space is compact iff every real valued continuous function achieves a max iff the range of every continuous real valued function is compact iff the nested intersection of every decreasing sequence of nonempty closed sets is nonempty.

In teaching the course on analysis I learned some nifty examples from a studentâs question. If you have continuous bijections X to Y and Y to X, are X and Y homeomorphic? Answer: no. Not even in R and not even for closed connected sets in R^2. Kuratowski, when he was a student of Sierpinski around 1920. But you probably knew these examples years ago. I did not. Whatâs great is when a beginning student asks a question to which you donât know the answer.

Best, Charles Pugh

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