Question

Suppose that \(\mathbf{F}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) is continuous and invertible on a compact set \(S .\) Show that \(\mathbf{F}_{S}^{-1}\) is continuous. HINT: If \(\mathbf{F}_{S}^{-1}\) is not continuous at \(\overline{\mathbf{U}}\) in \(\mathbf{F}(S),\) then there is an \(\epsilon_{0}>0\) and a sequence \(\left\\{\mathbf{U}_{k}\right\\}\) in \(\mathbf{F}(S)\) such that \(\lim _{k \rightarrow \infty} \mathbf{U}_{k}=\overline{\mathbf{U}}\) while \(\mid \mathbf{F}_{S}^{-1}\left(\mathbf{U}_{k}\right)-\mathbf{F}_{S}^{-1}\left(\overline{\mathbf{U})} \mid \geq \epsilon_{0}, \quad k \geq 1\right.\) Use Exercise 5.1 .32 to obtain a contradiction.

Short answer

Question: Show that if a continuous and invertible function F: R^n → R^n is defined on a compact set S, then its inverse F(S)^(-1) is continuous. Answer: We argue by contradiction. First, we assume that F(S)^(-1) is not continuous at a point 𝑈̅ in F(S). From this assumption, we can find an ε₀ > 0 and a sequence {𝑈_k} such that |F(S)^(-1)(𝑈_k) − F(S)^(-1)(𝑈̅)| ≥ ε₀ for all k. Next, we define sequence {𝑉_k} as 𝑉_k = F(S)^(-1)(𝑈_k) and apply Exercise 5.1.32 to find a subsequential limit 𝑉, concluding that lim(s→∞) 𝑉_(k_s) = 𝑉 . By the continuity of F, we have that lim(s→∞) F(𝑉_(k_s)) = F(𝑉). Consequently, we can deduce that lim(s→∞) F(S)^(-1)(𝑈_(k_s)) = 𝑉. However, this contradicts our initial assumption of |F(S)^(-1)(𝑈_k) − F(S)^(-1)(𝑈̅)| ≥ ε₀ . Therefore, F(S)^(-1) must be continuous.

Step-by-step solution

Assume that \(\mathbf{F}_{S}^{-1}\) is not continuous at \(\overline{\mathbf{U}}\)

According to the hint, we assume that \(\mathbf{F}_{S}^{-1}\) is not continuous at the point \(\overline{\mathbf{U}}\) in \(\mathbf{F}(S)\). Consequently, there exists an \(\epsilon_{0}>0\) and a sequence \(\left\\{ \mathbf{U}_k \right\\}\) in \(\mathbf{F}(S)\) such that: $$\lim_{k \rightarrow \infty} \mathbf{U}_k = \overline{\mathbf{U}}$$ and $$\left| \mathbf{F}_S^{-1}\left( \mathbf{U}_k \right) - \mathbf{F}_S^{-1}\left( \overline{\mathbf{U}} \right) \right| \geq \epsilon_{0}, \quad k \geq 1.$$

Define sequence \(\left\\{ \mathbf{V}_k \right\\}\) and apply Exercise 5.1.32

Let \(\mathbf{V}_k = \mathbf{F}_S^{-1}\left( \mathbf{U}_k \right)\). By exercising 5.1.32, since \(\mathbf{F}\) is continuous and invertible on the compact set \(S\), there exists a subsequence \(\left\\{ \mathbf{V}_{k_s} \right\\}\) of \(\left\\{ \mathbf{V}_k \right\\}\) such that $$\lim_{s \rightarrow \infty} \mathbf{V}_{k_s} = \mathbf{V},$$ where \(\mathbf{V} = \mathbf{F}_S^{-1}(\overline{\mathbf{U}})\).

Derive a contradiction

From Step 2, we know that \(\lim_{s \rightarrow \infty} \mathbf{V}_{k_s} = \mathbf{V}\). By applying the continuity of \(\mathbf{F}\), we have: $$\lim_{s \rightarrow \infty} \mathbf{F}\left(\mathbf{V}_{k_s}\right) = \mathbf{F}(\mathbf{V}),$$ which implies that $$\lim_{s \rightarrow \infty} \mathbf{U}_{k_s} = \overline{\mathbf{U}},$$ since \(\mathbf{U}_{k_s} = \mathbf{F}(\mathbf{V}_{k_s})\) and \(\overline{\mathbf{U}} = \mathbf{F}(\mathbf{V})\). This implies that \(\lim_{s \rightarrow \infty} \mathbf{F}_S^{-1}\left( \mathbf{U}_{k_s} \right) = \mathbf{V}\). But by our initial assumption, we have $$\left| \mathbf{F}_S^{-1}\left( \mathbf{U}_k \right) - \mathbf{F}_S^{-1}\left( \overline{\mathbf{U}} \right) \right| \geq \epsilon_{0}.$$ This is a contradiction, since \(\lim_{s \rightarrow \infty} \mathbf{F}_S^{-1}\left( \mathbf{U}_{k_s} \right) = \mathbf{V}\). Therefore, our initial assumption that \(\mathbf{F}_S^{-1}\) is not continuous must be incorrect.

Conclude that \(\mathbf{F}_S^{-1}\) is continuous

As we have derived a contradiction from the assumption that \(\mathbf{F}_S^{-1}\) is not continuous at \(\overline{\mathbf{U}}\), we can conclude that \(\mathbf{F}_{S}^{-1}\) is indeed continuous.

Key concepts

Compact Sets in Real Analysis

In real analysis, a compact set is a fundamental concept that plays a crucial role in various aspects of mathematics, including the study of continuous functions. A set is considered compact if it is both closed (contains all its limit points) and bounded (all its points lie within a certain distance from each other).

The most famous result concerning compact sets in real analysis is perhaps the Heine-Borel theorem, which states that a subset of \(\mathbb{R}^n\) is compact if and only if it is closed and bounded. Compactness is significant because it allows the use of the Bolzano-Weierstrass theorem, ensuring every sequence in the compact set has a convergent subsequence within the set.

When analyzing the continuity of inverse functions, compactness becomes essential. If a function is continuous on a compact set, then the image of this set under the function is also compact. This property is leveraged to show the continuity of the inverse function, as seen in the original exercise provided.

Sequences and Limits in Real Analysis

Sequences and their limits form the baseline for understanding topics such as continuity, convergence, and much more in real analysis. A sequence in \(\mathbb{R}^n\) is an ordered set of points listed one after another, and the concept of a limit pertains to the behavior of a sequence as it progresses towards infinity.

The limit of a sequence is the value that the sequence's terms get arbitrarily close to as the index increases indefinitely. If a sequence converges to some limit \( L \), we can write \(\lim_{k \rightarrow \infty} \mathbf{U}_k = L\). In the practice of real analysis, establishing the limit of a sequence is vital in proving continuity, as seen in the solution steps for the continuity of the inverse function. Applying the definition of limits, if the sequence formed by the images \(\mathbf{U}_k\) of the inverse function \(\mathbf{F}_{S}^{-1}\) converges to \(\overline{\mathbf{U}}\), then for \(\mathbf{F}_{S}^{-1}\) to be continuous, the sequence \(\mathbf{F}_{S}^{-1}(\mathbf{U}_k)\) must converge to \(\mathbf{F}_{S}^{-1}(\overline{\mathbf{U}})\).

Contradiction in Mathematical Proofs

Contradiction is a powerful and often used technique in mathematical proofs. The method of contradiction starts by assuming the opposite of what we are trying to prove. If this assumption leads to a logical inconsistency, then the initial assumption must be false, and hence, the statement we are trying to prove is true.

In the context of the given exercise, the proof utilizes contradiction by assuming that the inverse function \(\mathbf{F}_{S}^{-1}\) is not continuous. It then proceeds to demonstrate that this leads to a situation that cannot possibly be true (a contradiction). Specifically, it shows that a certain sequence does not converge in the manner it should if the inverse were truly discontinuous. This contradiction indicates that our original assumption - the discontinuity of \(\mathbf{F}_{S}^{-1}\) - is incorrect, and thereby, it must be continuous.

The beauty of contradiction lies in its ability to navigate complex proofs by simplifying them into manageable scenarios where the false assumptions clearly fail to hold thus, strengthening the truth of the assertion being proven.