# Class of functions that map certain spaces to totally bounded spaces?

Suppose $$f:X\rightarrow Y$$ is an onto function where $$X$$ is a topological space and $$Y$$ is a metric space. For $$Y$$ to be totally bounded, I pick a sequence $$(y_n)$$ from $$Y$$ and try to show it has a Cauchy subsequence. Now $$(y_n)=(f(x_n))$$ for a certain sequence $$(x_n)$$ in $$X$$. If $$X$$ were sequentially compact, $$(x_n)$$ would have a convergent subsequence $$(x_{n_k})$$. Then, if $$f$$ were continuous, $$(y_{n_k})$$ would be convergent, hence Cauchy. If $$X$$ were totally bounded metric space, then $$(x_n)$$ would have a Cauchy subsequence $$(x_{n_k})$$. Then, if $$f$$ were to be uniformly continuous, $$(y_{n_k})$$ would be Cauchy. And if $$X$$ is compact, then $$Y$$ being its continuous image is compact as well. What other conditions can we impose on $$X$$ and/or $$f$$ so that $$Y$$ is totally bounded or compact?

• A sequence of points of a compact space can fail to have a convergent subsequence. For instance, the Čech-Stone compactification of natural numbers endowed with discrete topology has only trivial convergent sequences [Eng, Corollary 3.6.15]. The property you refer to concerns sequentially compact spaces. A metrizable space is compact iff it is sequentially compact. Each sequentially compact space is feebly compact. – Alex Ravsky Aug 13 at 21:11
• @AlexRavsky Yes, sorry. I had edited parts of the question. Earlier I said $X$ was a metric space. I'll edit the question. – Hrit Roy Aug 13 at 21:16

I don’t know natural conditions assuring total boundedness of $$Y$$ when $$f$$ is discontinuous. When $$f$$ is continuous, I think a rather wide and natural sufficient condition imposed on $$X$$ is its functional boundedness, that is, each real-valued function $$g$$ on $$X$$ is bounded. We can show this as follows. Recall that a space is feebly compact, if each locally finite family of non-empty open subsets of the space $$X$$ is finite. It is easy to show that both functionally bounded and feebly compact spaces are preserved by continuous maps onto. Also it is well-known (see Theorem 3.10.22 from [Eng]), that a Tychonoff space $$X$$ is functionally bounded iff it is feebly compact. On the other hand, there exist feebly compact non-regular spaces (for instance, the segment $$[0,1]$$, where the usual topology was strengthened by declaring a set $$[0,1]\setminus\{1/n:n\in\Bbb N\}$$ open). Since each metric space is Tychonoff, each feebly compact space is functionally bounded. Finally, each functionally bounded subset of a metric space is feebly compact, so countably compact by Theorem 3.10.21 from [Eng] and compact by Theorem 5.1.20 from [Eng] (because each metric space is paracompact by The Stone Theorem (Theorem 4.4.1 from [Eng])).

References

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

Theorem 3.10.22 For every Tychonoff space $$X$$ the following conditions are equivalent:

(i) The space X is pseudocompact.

(ii) Every locally finite family of non-empty open subsets of $$X$$ is finite.

(iii) Every locally finite open cover of $$X$$ consisting of non-empty sets is finite.

(iv) Every locally finite open cover of $$X$$ has a finite subcover.

Proof. First we shall show that (i)$$\Rightarrow$$ (ii). Suppose that (ii) does not hold; thus there exists a locally finite family $$\{U_i\}_{i=1}^\infty$$ of non-empty open subsets of $$X$$. Let us choose a point $$x_i\in U_i$$ for $$i=1, 2, \dots$$ Since $$X$$ is a Tychonoff space, for $$i=1, 2, \dots$$ there exists a continuous function $$f_i: X\to\Bbb R$$ such that $$f(x_i)=i$$ and $$f_i(X\setminus U_i)\subset \{0\}$$. From the local finiteness of the family $$\{U_i\}_{i=1}^\infty$$ it follows that the formula $$f(x) = \sum_{i=1}^\infty |f_i(x)|$$ defines a continuous function $$f:X\to \Bbb R$$; as $$f$$ is not bounded, the space $$X$$ is not pseudocompact.

The implications (ii)$$\Rightarrow$$(iii) and (iii)$$\Rightarrow$$(iv) are obvious; to conclude the proof it suffices to show that (iv)$$\Rightarrow$$(i). Let $$f$$ be a continuous real-valued function defined on a space $$X$$ satisfying (iv). Clearly, the family $$\{f^{-1}((i — 1, i + 1)) : i = 0, \pm 1, \pm 2,\dots \}$$ is a locally finite open cover of $$X$$; the existence of a finite subcover implies that $$f$$ is bounded.$$\square$$

Theorem 3.10.21. Every pseudocompact normal space is countably compact.

Proof Assume that $$X$$ is a normal space which is not countably compact. Thus there exists a set $$A = \{x_1, x_2,\dots\}\subset X$$ such that $$x_i\ne x_j$$, whenever $$i\ne j$$ and $$A^d=\varnothing$$. Clearly $$A$$ is a discrete closed subspace of $$X$$ and by the Tietze-Urysohn theorem there exists a continuous function $$f:X\to\Bbb R$$ such that $$f(x_i)=i$$ for $$i= 1,2,\dots$$ Since $$f$$ is not bounded, the space $$X$$ is not pseudocompact. $$\square$$

Theorem 5.1.20. Every countable compact paracompact space is compact.

Proof. Let $$\mathcal A$$ be an open cover of a countably compact paracompact space $$X$$. It follows from Theorem 3.10.3 that any locally finite open refinement $$\mathcal B$$ of $$A$$ is finite, so that the space $$X$$ is compact.$$\square$$

• Interesting and helpful. Pseudocompactness would also work right? – Hrit Roy Aug 12 at 15:31
• @HritRoy Right, pseudocompact spaces are exactly Tychonoff feebly compact spaces. – Alex Ravsky Aug 12 at 18:40
• I thought feebly compact implies pseudocompact, but not necessarily the other way? If I know that it's true for pseudocompact spaces, it's anyway true for feebly compact spaces then. – Hrit Roy Aug 13 at 8:59
• @HritRoy Each pseudocompact space is feebly compact (by Theorem 3.10.22). But pseudocompact spaces are Tychonoff by definitions, whereas feebly compact spaces can be non-Tychonoff. – Alex Ravsky Aug 13 at 21:01
• I'm not assuming $X$ to be Tychonoff. – Hrit Roy Aug 13 at 21:17