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?

  • $\begingroup$ 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. $\endgroup$ Aug 13, 2020 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, 2020 at 21:16

1 Answer 1


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])).


[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.$\square$

  • 1
    $\begingroup$ Interesting and helpful. Pseudocompactness would also work right? $\endgroup$
    – Hrit Roy
    Aug 12, 2020 at 15:31
  • $\begingroup$ @HritRoy Right, pseudocompact spaces are exactly Tychonoff feebly compact spaces. $\endgroup$ Aug 12, 2020 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, 2020 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. $\endgroup$ Aug 13, 2020 at 21:01
  • $\begingroup$ I'm not assuming $X$ to be Tychonoff. $\endgroup$
    – Hrit Roy
    Aug 13, 2020 at 21:17

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