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Let \(f: R \rightarrow R\) be defined by $$ \begin{aligned} &f(x)=\frac{1}{x}, x>0, \\ &f(x)=0, x \leqq 0 . \end{aligned} $$ Prove that the graph \(\Gamma_{f}\) is a closed subset of \(\left(R^{2}, d\right)\), but that \(f\) is not continuous.

Short Answer

Expert verified
\(\Gamma_f\) is closed, but \(f\) is discontinuous at \(x=0\).

Step by step solution

01

Understand the Graph of the Function

The graph \(\Gamma_f\) consists of two parts: (1) the set \( \{ (x, \frac{1}{x}) \mid x > 0 \} \), which is a hyperbola like curve in the \(xy\)-plane, and (2) the set \( \{ (x, 0) \mid x \leq 0 \} \), which is the non-positive side of the \(x\)-axis.
02

Prove the Graph is Closed

A set is closed if its complement is open. Consider any sequence of points \((x_n, f(x_n)) \to (a, f(a))\) in \(R^2\). If \(x_n > 0\), then \(f(x_n) = \frac{1}{x_n}\). In this case, as \(x_n \to a > 0\), \(f(x_n) \to \frac{1}{a}\). If \(x_n \leq 0\), \(f(x_n)\) is always 0. Hence, for any sequence, the limits conform to the graph, showing \(\Gamma_f\) is a closed set.
03

Prove Non-Continuity of \(f\)

For a function to be continuous at point \(a\), \(\lim_{x \to a} f(x) = f(a)\). For \(a=0\), from the right side (\(x\to 0^+\)), \(f(x)=\frac{1}{x}\to \infty\). Since \(f(0) = 0\), \(\lim_{x \to 0} f(x) eq f(0)\), indicating discontinuity at \(x=0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closed Set
In topology, a set is termed "closed" if its complementary set is open. To understand what this means, consider the set we're evaluating: the graph of the function \( \Gamma_f \). For our function \( f \), the graph is composed of two distinct parts: the hyperbola-like curve \( \{ (x, \frac{1}{x}) \mid x > 0 \} \), and the line segment \( \{ (x, 0) \mid x \leq 0 \} \) along the non-positive part of the x-axis.

The critical aspect of proving that this graph is a closed set involves looking at what happens to sequences of points within this set. If any sequence of points \( (x_n, f(x_n)) \) within the graph converges to a point \( (a, f(a)) \) in \( \mathbb{R}^2 \), that limit point must also lie on the graph.
  • When \( x_n \gt 0 \), the function is \( f(x_n) = \frac{1}{x_n} \), and as \( x_n \to a \gt 0 \), then \( f(x_n) \to \frac{1}{a} \).
  • If \( x_n \leq 0 \), the function is \( f(x_n) = 0 \).
These convergences demonstrate that both limit points are part of the graph, affirming \( \Gamma_f \) as a closed set.

Finding the complement (everything not in \( \Gamma_f \)) that remains open in the plane hence helps prove the graph is closed.
Continuity
Continuity of a function involves the behavior of the function as it approaches a point from all directions. For a function \( f \) to be continuous at a point \( a \), the limit as \( x \) approaches \( a \) must equal \( f(a) \). In simpler terms, \( \lim_{x \to a} f(x) = f(a) \). For our function, let's focus particularly at the point \( x = 0 \). Here, the function is defined as \( f(x) = 0 \) for \( x \leq 0 \). However, notice that approaching zero from the right (i.e., \( x \to 0^+ \)), the function \( f(x) = \frac{1}{x} \) tends towards infinity.This behavior tells us something crucial:
  • Approaching from the right results in \( \lim_{x \to 0^+} f(x) = \infty \).
  • Approaching from the left (\( x \leq 0 \)), \( \lim_{x \to 0^-} f(x) = 0 \).
  • Since \( f(0) = 0 \), the limits do not match: \( \lim_{x \to 0} f(x) eq f(0) \).
This discrepancy indicates a discontinuity at \( x = 0 \), and thus, \( f \) is not continuous at this point.
Graph of a Function
The graph of a function is a visual representation of the set of all ordered pairs \( (x, f(x)) \). For the given function \( f \), the graph is partitioned into two segments:

  • A hyperbolic curve \( \{ (x, \frac{1}{x}) \mid x > 0 \} \) in the first quadrant, which corresponds to parts of the curve where \( x \) is positive.
  • The x-axis for \( \{ (x, 0) \mid x \leq 0 \} \), representing all negative and zero values where the function constantly outputs zero.
Understanding these parts helps in visualizing how the function behaves on different domains.

When faced with such graphs, one should pay attention to:
  • The domain restrictions, which provide context for each section of the graph.
  • The possible asymptotic behavior, like the approach towards \( x = 0 \) from both directions.
  • The junction points where different forms of the function meet, as these often indicate points of interest related to continuity and limit behavior.
By examining such features, the behavior of the function can be predicted and understood even without explicit calculations, offering a powerful tool in mathematical analysis.

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Most popular questions from this chapter

A sequence of real numbers \(a_{1}, a_{2}, \ldots\) is called monotone non- decreasing if \(a_{i} \leqq a_{i+1}\), for each \(i\) and called monotone non- increasing if \(a_{i} \geqq a_{i+1}\), for each \(i\). A sequence which is either monotone non-decreasing or monotone non-increasing is said to be monotone. The sequence is said to be bounded above if there is a number \(K\) such that \(a_{i} \leqq K\) for each \(i\) and bounded below if there is a number \(M\) such that \(a_{i} \geqq M\) for each \(i\). A sequence which is both bounded above and bounded below is called bounded. Prove that a convergent sequence of real numbers is bounded. Prove that a monotone non-decreasing sequence of real numbers which is bounded above converges to a limit \(a\) and that \(a\) is the l.u.b. of the set \(\left\\{a_{1}, a_{2}, \ldots\right\\}\). Similarly prove that a monotone non-increasing sequence which is bounded below converges to a limit \(b\) and that \(b\) is the g.l.b. of the set \(\left\\{a_{1}\right.\), \(\left.a_{2}, \ldots\right\\}\)

Let \(X\) be the set of continuous functions \(f:[a, b] \rightarrow R .\) Let \(d^{*}\) be the distance function on \(X\) defined by $$ d^{*}(f, g)=\int_{a}^{b}|f(t)-g(t)| d t, $$ for \(g \in X\). For each \(f \in X\), set $$ I(f)=\int_{a}^{b} f(t) d t $$ Prove that the function \(I:\left(X, d^{*}\right) \rightarrow(R, d)\) is continuous.

Let \(\left(Y, d^{\prime}\right)\) be a subspace of the metric space \((X, d)\). Prove that a subset \(O^{\prime} \subset Y\) is an open subset of \(\left(Y, d^{\prime}\right)\) if and only if there is an open subset \(O\) of \((X, d)\) such that \(O^{\prime}=Y \cap O\). Prove that a subset \(F^{\prime} \subset Y\) is a closed subset of \(\left(Y, d^{\prime}\right)\) if and only if there is a closed subset \(F\) of \((X, d)\) such that \(F^{\prime}=Y\) \(\cap F\). For a point \(a \in Y\), prove that a subset \(N^{\prime} \subset Y\) is a neighborhood of \(a\) if and only if there is a neighborhood \(N\) of \(a\) in \((X, d)\) such that \(N^{\prime}=Y \cap N\).

Let \(f: X \rightarrow Y\) be a function from a metric space \(X\) into a metric space \(Y\). Let \(a \in X\) and let \(B_{f(a)}\) be a basis for the neighborhood system at \(f(a)\). Prove that \(f\) is continuous at \(a\) if and only if for each \(N \in \mathbb{B}_{f(a)}, f^{-1}(N)\) is a neighborhood of \(a\).

Consider the subspace \(\left(Q, d_{Q}\right)\) (the rational numbers) of \((R, d) .\) Let \(a_{1}, a_{2}\), \(\ldots\) be a sequence of rational numbers such that \(\lim _{n} a_{n}=\sqrt{2}\). Prove that, given \(\varepsilon>0\), there is a positive integer \(N\) such that for \(n, m>N,\left|a_{n}-a_{m}\right|<\) \(\varepsilon\). Does the sequence \(a_{1}, a_{2}, \ldots\) converge when considered to be a sequence of points of \(\left(Q, d_{Q}\right)\) ?

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