4
$\begingroup$

Let $(X,\mathfrak{T})$ be a topological space. We know that if $X$ is compact and $f:X\to \mathbb{R}$ be any continuous function then $f(X)$ is bounded since the continuous image of a compact set is compact and any compact subset of a metric space is bounded.

My questions are (edited after David C. Ullrich's comment below),

  • What is(are) the necessary and/or sufficient condition(s) on $(X,\mathfrak{T})$ so that if for any continuous $f:X\to \mathbb{R}$ we can conclude that $f(X)$ is bounded then it would imply that $X$ is compact?

  • Let $(X,\mathfrak{T})$ be a topological space. If every continuous function $f:X\to Y$ is bounded for all metric space $(Y,d)$ then can we say that $X$ is compact?

  • Let $(X,\mathfrak{T})$ be a topological space. If there exists a metric space $(Y,d)$ such that every continuous function $f:X\to Y$ is bounded then can we say that $X$ is compact?

$\endgroup$
  • $\begingroup$ The first uncountable ordinal is a counterexample. $\endgroup$ – David C. Ullrich May 8 '16 at 15:08
  • $\begingroup$ This is not true in general for topological spaces. However, this is true for metric spaces. $\endgroup$ – user175531 May 8 '16 at 15:10
2
$\begingroup$

A space $X$ is called pseudocompact when every real-valued continuous function on $X$ is bounded. This is a well-studied notion. Indeed a compact space, or even a countably compact space (every countable open cover has a finite subcover) is pseudocompact.

A classical theorem: for normal and $T_1$ spaces, countably compact and pseudocompact are equivalent.

The classical example of $\omega_1$, the first uncountable ordinal, is pseudocompact (and countably compact) and not compact. This shows that we cannot make the jump from countably compact to compact.

We can use Lindelöf (every cover has a countable subcover) as the gap property to compactness. So if we use at least Tychonoff spaces, a class of spaces where pseudocompactness and compactness coincide are the Lindelöf spaces.

Another class are the metric spaces, because for metric spaces countable compactness, pseudocompactness and compactness are equivalent.

A pseudocompact space has the property that every continuous map into every metric space is bounded. This follows from the equivalence for metric spaces and the fact that the continuous image of a pseudocompact space is pseudocompact.

So $\omega_1$ is also a counterexample to question 2, and a fortiori to 3 as well.

$\endgroup$
1
$\begingroup$

For the second question...

You consider $\mathbb{R}_d$ with discrete topology and $[0,1]$ with euclidean topology. If $f:\mathbb{R}_d\longrightarrow[0,1]$ is a function, then $f$ is continuos and bounded, but $\mathbb{R}_d$ is not compact.

$\endgroup$
  • $\begingroup$ $Y$ is any metric space, i.e., $Y$ varies over the class of all metric spaces. See the edit. I hope that is clearer now. $\endgroup$ – user 170039 May 8 '16 at 15:34
  • $\begingroup$ Yes...Now the problem is more difficult! xD $\endgroup$ – Vincenzo Zaccaro May 8 '16 at 15:37
1
$\begingroup$

The first uncountable ordinal $X = \omega_1$, with its order topology, is a counterexample for the second question (and hence also the third). It's well known that $X$ is not compact, yet any continuous function $h : X \to \mathbb{R}$ is bounded (in fact, it's eventually constant). Now if $(Y,d)$ is any metric space and $f : X \to Y$ is continuous, fix a point $y_0 \in Y$ and let $g : Y \to \mathbb{R}$ be given by $g(y) = d(y,y_0)$, which is continuous. Then $h = g \circ f$ is a continuous map from $X$ to $\mathbb{R}$, hence bounded; say $h(x) \le M$ for all $x \in X$. That means $d(y_0, f(x)) \le M$ for all $x \in X$, so that $f(X)$ is bounded.

$\endgroup$
-1
$\begingroup$

Let's say a space is RB if every continuous real-valued function is bounded.

You ask for necessary and sufficient conditions such that RB implies compact. The word "implies" is a little funny; what people often mean by it is difficult if not impossible to define precisely.

The only precise definition of "A implies B" that I know is "A is false or B is true". If we take that as the definition of "implies" then the answer to your question is that RB implies compact if and only if $X$ is compact or there exists an unbounded real-valued continuous function on $X$.

Almost surely not an answer to what you really wanted to know, but possibly the best answer that question is going to get. (The best answer that actually gives a necessary and sufficient condition.)

$\endgroup$
  • $\begingroup$ I don't understand what you meant by "The only precise definition of "A implies B" that I know is "A is false or B is true"." Can you clarify? $\endgroup$ – user 170039 May 8 '16 at 17:13
  • $\begingroup$ @user170039 I don't understand what you want me to clarify. That is the only precise definition of "implies" that I know... $\endgroup$ – David C. Ullrich May 8 '16 at 17:18
  • $\begingroup$ Whoever downvoted this should give a better answer to the first question. $\endgroup$ – David C. Ullrich May 8 '16 at 17:19
  • $\begingroup$ Why do you think that it is precise? For me this is precise enough. $\endgroup$ – user 170039 May 8 '16 at 17:20
  • 1
    $\begingroup$ It's surprising that you spent time to write this obviously unhelpful answer (which I think would be better left as a comment). By the way, I don't see anywhere mentioning that your definition of implies is not precise. I only wanted you to clarify the reason for which you think that your definition was precise. $\endgroup$ – user 170039 May 9 '16 at 12:48

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.