The Boundedness Theorem and the Extreme Value Theorem for subsets of finite-dimensional (Euclidean) space

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In an earlier post we looked at the Boundedness Theorem and the Extreme Value Theorem for continuous, real-valued functions on closed and bounded intervals $[a,b]$. However, the proof we gave there generalises to any setting in which the lemma we used is valid.

Let $D$ be a non-empty subset of $\R^2$, and consider the following property (B) (non-standard!) that $D$ may or may not have:

(B)        Every continuous function $f:D\to\R$ is bounded above on $D$.

By the reasoning in the earlier post, we can show that the Extreme Value Theorem holds for any non-empty set $D$ that satisfies condition (B). So the question arises, which subsets of $\R^2$ satisfy condition (B)?

By considering the continuous function $f(x,y)=\sqrt{x^2+y^2}$ (or we could use $f(x,y)=x^2+y^2$), it is clear that any subset $D$ of $\R^2$ that satisfies condition (B) must be a bounded subset of $\R^2$ (that is, the set of distances from points of $D$ to $(0,0)$ is bounded above). But this is not enough, because e.g. if $D=\{(x,y)\in\R^2:x^2+y^2<1\}$ (a disc without its boundary), then the continuous function $f:D\to\R$ defined by $f(x,y)=1/(1-x)$ is not bounded above on $D$.

It turns out that what we need here (and more generally for subsets of $\R^d$ for any positive integer $d$) is for the non-empty set $D$ to be closed and bounded. But where boundedness of sets is relatively easy to explain (in terms of distances to the origin, as above for $\R^2$), it is perhaps less common to try to explain to first-year students what closed means.

Rather than proving the more general fact here, let me focus on proving that closed and bounded rectangles in $\R^2$ (i.e., sets of the form $[a,b]\times [c,d]$) satisfy condition (B). (This has obvious generalisations to higher dimensions.)

Theorem

Let $a,b,c,d \in \R$ with $a<b$ and $c<d$, and set $D=[a,b]\times [c,d] \subseteq \R^2$.

Let $f:D\to\R$ be a continuous function. Then $f$ is bounded above on $D$.

Thus $D$ has property $(B)$.

Proof

Assume, towards a contradiction, that $f$ is not bounded above on $D$. Then, in particular, for each $n\in\N$, we can find a point $(x_n,y_n)\in D$ such that $f(x_n,y_n)>n$. (This is because $n$ is not an upper bound for $f(D)$.)

[The idea is now to take a convergent subsequence of our sequence $(x_n,y_n)$ of elements of $D$. However, at this point we may only know the Bolzano-Weierstrass Theorem for $\R$ and not for $\R^2$. So we have a choice: establish the Bolzano-Weierstrass Theorem in $\R^2$, or establish a special case here for sequences in $[a,b]\times [c,d]$. We also need to explain why the limit of any such convergent subsequence must still be in $D$, but this is relatively easy for this kind of set $D=[a,b]\times [c,d]$. I'll leave these details for a separate post! Either way ...]

There must exist a point $(x_0,y_0)\in D$ and positive integers $n_1<n_2<n_3<\cdots$ such that $x_{n_k}\to x_0$ and $y_{n_k} \to y_0$ as $k\to \infty$, i.e., $(x_{n_k},y_{n_k})\to(x_0,y_0)$ as $k\to\infty$. Since $f$ is continuous on $D$, we should have $$f(x_{n_k},y_{n_k}) \to f(x_0,y_0) \in\R \text{ as } k \to \infty\,.$$ But in fact we have $$f(x_{n_k},y_{n_k}) > n_k \to \infty \text{ as } k\to \infty\,,$$ and so we have obtained a contradiction. This shows that $f$ must be bounded above on $D$.

Since this holds for all continuous, real-valued functions on $D$, $D$ has property (B), as claimed.$\quad\square$

Corollary

The Boundedness Theorem and the Extreme Value Theorem hold for sets of the form $D=[a,b]\times [c,d]\subseteq \R^2$.

Proof

Repeat for $D$ the proof given in the 1d setting from the earlier post. $\quad\square$

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