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I have the following definition of bounded subset in a metric space:

Let $X\subset M$ where $M$ is metric space. Then $X$ is said to be bounded if for all $x,y\in X, d(x,y) \le c$ for some $c$.

And I have these two definitions of a bounded function $f: X\to S$ where $X$ is arbitrary set and $S$ is a vector space with a norm $|\cdot|$ (made metric space through $d(x,y) := |x-y|$):

1) $f$ is said to be bounded if $f(X)$ is a bounded subset of $S$. 2) $f$ is said to be bounded if $|f(x)| \le k$ for some k.

I want to see whether these definitions are equivalent. 1 follows easily from 2 using triangle inequality:

$$|f(x) - f(y)| \le |f(x)|+|f(y)| \le 2c. $$

But I'm stuck in the converse. I think one problem is that the $0$ vector of $S$ is not necessarily in the image $f(X)$ but I'm still getting nothing from this. Any help will be appreciated.

AnalyticHarmony
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2 Answers2

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The equivalence will become obvious if you prove the following: when the metric space $M$ is actually a normed vector space , boundedness of a subset $X$ is equivalent to the fact that $\sup \{|x|:x\in X\} <\infty$. To prove the equivalence argue as follows: if $C\equiv \sup \{|x|:x\in X\} <\infty$ then $d(x,y)=|x-y| \leq 2C$ for all $x,y \in X$. Conversely if $d(x,y) \leq C$ for all $x,y \in X$ fix a point $x_0 \in K$ and use triangle inequality to get $|x| \leq |x-x_0|+|x_0| \leq C+|x_0|$ for all $x \in X$ proving that $\sup \{|x|:x\in X\} <\infty$.

Kavi Rama Murthy
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For the converse, assume that $f(X)$ is bounded: there is a constant $c>0$ such that for any $y,y'\in f(X)$, we have $|y-y'|<c$.

Fix $x_0\in X$. Let $x\in X$ be any element of $X$. We have

$$|f(x)-f(x_0)|<c$$

On the other hand, by the reverse triangle inequality, we have

$$|f(x)|-|f(x_0)|\leq \bigl||f(x)|-|f(x_0)|\bigr|\leq|f(x)-f(x_0)|$$

It follows that $$|f(x)|<x+|f(x_0)|$$

This result being true for every $x\in X$, taking the right hand side to be $k$, we see that $f$ is a bounded function in the sense of $2)$.

Suzet
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