If $A$ is a connected subset of a real normed linear space such that every real valued continuous function on $A$ is uniformly continuous , then is it true that $A$ is bounded ? If not , then what if we restrict $A$ to be path connected ? Even if not in this case , does convex property on $A$ guarantee boundness ? Do we require the real normed linear space to be finite dimensional ? ( The only thing I know is that if $X$ is a metric space such that every real valued continuous function on $X$ is uniformly continuous , then $X$ is complete )
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2 - $\begingroup$ With smile: have you considered one-question-at-one-time? $\endgroup$Yes– Yes2015-10-04 07:25:01 +00:00Commented Oct 4, 2015 at 7:25
- $\begingroup$ If our $A$ contains a ray (i.e. a set of the form $\{v+wt : t\ge 0\}$) then we can find a real-valued continuous, but not uniformly continuous function on $A$ - we just extend a non-uniformly-continuous function over a ray.) I suspect that such method can apply on an unbounded convex set. $\endgroup$Hanul Jeon– Hanul Jeon2015-10-04 07:38:32 +00:00Commented Oct 4, 2015 at 7:38
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1 Answer
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If $X$ is a connected metric space such that every real valued continuous function on $X$ is uniformly continuous, then $X$ is compact. In particular, it is bounded.
Proof: If $X$ is not compact, there is a sequence $S = \{x_n\}_{n=1}^\infty$ with no convergent subsequence. Then for each $n$ there is $\epsilon_n > 0$ such that the ball $B_{\epsilon_n}(x_n)$ contains no members of $S$ except $x_n$. We may assume $\epsilon_n \to 0$ as $n \to \infty$. Define a function $f$ on $X$ by:
$$f(x) = \sup(0, \sup_n (1 - 2 d(x, x_n)/\epsilon_n))$$
Note that for any $x$ there is at most one $x_n$ with $2 d(x,x_n) < \epsilon_n$. Then $f$ is continuous, but not uniformly continuous. In fact, $f(x_n) = 1$ but $f(x) = 0$ if $d(x,x_n) = \epsilon_n/2$ (and by connectedness, if $\epsilon_n/2$ is small enough there is such an $x$).
