Question about boundedness of a continuous function $f$ without using open cover and its finite subcover.

Question

Let $f:I=[a,b] \to \mathbb{R}$ be a continuous function. It is given that for any $x \in I$ there exists a neighbourhood $V_\delta(x)$ of $x$ on which $f$ is bounded. Then need to show that $f$ is bounded on $I$.

If I use $I$ is a compact subset of $\mathbb{R}$ and use the open cover definition of compact set the I am able to solve this. But is it possible to solve the above question without using the open cover thing, in particular using real analysis?

I tried and was unable to solve this. Thanks for your concern.

Answer 1

Here is an answer that hides all the topological work. If $f$ is continuous and unbounded so is $|f|$, so assume WLOG that $f$ maps into the non-negative numbers. Then $\tau: [0,1] \to \mathbb R$ defined by $x \mapsto \sup \{ f(y) : y \in [0,x] \}$ is also continuous (this is not too difficult to check) and $\tau(x) \to \infty$ as $x \to 1$, a contradiction.

Answer 2

It follows from the completeness axiom:

Given the closed interval $[a,b]$ and a positive valued function $\delta\colon[a,b]\to (0,\infty)$, a $\delta$-fine tagged partition of $[a,b]$ is a set $$^t\!P_\delta=\{([x_{i-1},x_i],t_i):1\le i\le n\}$$ where $x_0=a$, $x_n=b$, $x_{i-1}<x_i$, and $t_i\in[x_{i-1},x_i]$ for all $1\le i\le n$. Furthermore, for all $1\le i\le n$, $[x_{i-1},x_i]\subseteq (t_i-\delta(t_i),t_i+\delta(t_i))$.

Theorem: For any interval $[a,b]$ and function $\delta\colon[a,b]\to (0,\infty)$, there exists a $\delta$-fine tagged partition of $[a,b]$.

Proof: Let $D$ be the set of all numbers $x$, where $a<x\le b$, for which there is a $\delta$-fine tagged partition of $[a,x]$. By taking $x=a+\delta(a)/2$, we see that $\{([a,a+\delta(a)/2],a)\}$ is a $\delta$-fine tagged partition of $[a,a+\delta(a)/2]$. Thus, $D$ is non-empty. Since $D$ is bounded above by $b$, it has a least upper bound --- call it $\beta$. We claim that $\beta\in D$ and $\beta=b$, and conclude that there is a $\delta$-fine tagged partition of $[a,b]$.

First we must show that $\beta\in D$. Since $\beta$ is the least upper bound of $D$, there is some $r\in(0,\delta(\beta)/2)$ such that we have a $\delta$-fine tagged partition of $[a,\beta-r]$. If necessary, we can remove tagged intervals or part of tagged intervals from this partition to obtain a $\delta$-fine tagged partition of $[a,\beta-\delta(\beta)/2]$. By adjoining the singleton $\{([\beta-\delta(\beta)/2,\beta],\beta)\}$, we have a $\delta$-fine tagged partition of $[a,\beta]$.

Now suppose that $\beta<b$, and let $^t\!P_\delta=\{([x_{i-1},x_i],t_i):1\le i\le n\}$ be a $\delta$-fine tagged partition of $[a,\beta]$. Then since $\delta(\beta)$ is positive, we can extend $^t\!P_\delta$ to the tagged partition $^t\!P_\delta \cup \{([\beta,\beta+\delta(\beta)/2],\beta)\}$. Since this is a $\delta$-fine tagged partition of $[a,\beta+\delta(\beta)/2)]$, we have a contradiction (since we said $\beta$ is an upper bound for $D$ and have shown that $\beta+\delta(\beta)/2\in D$). Thus $\beta=b$, as desired.

Now let $\delta(x)$ be any positive value of $x$ such that $f$ is bounded on $(x-\delta(x),x+\delta(x))$. The previous theorem gives a $\delta$-fine tagged partition which by definition is finite.