3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alex Ravsky Aug 13 at 21:11
  • $\begingroup$ @AlexRavsky Yes, sorry. I had edited parts of the question. Earlier I said $X$ was a metric space. I'll edit the question. $\endgroup$ – Hrit Roy Aug 13 at 21:16
1
+50
$\begingroup$

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.$\square$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Interesting and helpful. Pseudocompactness would also work right? $\endgroup$ – Hrit Roy Aug 12 at 15:31
  • $\begingroup$ @HritRoy Right, pseudocompact spaces are exactly Tychonoff feebly compact spaces. $\endgroup$ – Alex Ravsky Aug 12 at 18:40
  • $\begingroup$ 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. $\endgroup$ – Hrit Roy Aug 13 at 8:59
  • $\begingroup$ @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. $\endgroup$ – Alex Ravsky Aug 13 at 21:01
  • $\begingroup$ I'm not assuming $X$ to be Tychonoff. $\endgroup$ – Hrit Roy Aug 13 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.