Cluster point, a, and bounded f, f(a) <= y

[G-X]

Let [Math]f : D[/Math]->[Math] R[/Math] where [Math]D \subset R^{k}[/Math], [Math]k = 1[/Math] or [Math]2[/Math]. Let f be continuous at [Math]a \in D[/Math]and let [Math]a[/Math] be a cluster point of a subset [Math]E[/Math] of [Math]D[/Math]. If there is a real number [Math]y[/Math] such that for each [Math]x \in E,[/Math] [Math]f(x) \le y[/Math], then [Math]f(a) \le y[/Math].

This is what I have so far:

Given: The range is in [Math]R^{1}[/Math] thus the domain is restricted to [Math]R^{1}[/Math].
Given: [Math]f[/Math] is continuous at [Math]a[/Math]Given: [Math]a[/Math] is a cluster point of [Math]E[/Math]Given: [Math]f[/Math] is bounded [Math]\forall x \in E[/Math]Given: [Math]a \in D[/Math]
Case 1: Since, [Math]E \subset D[/Math], let [Math]a \in E[/Math]: Since [Math]f[/Math] is bounded [Math]\forall x \in E[/Math] such that

[Math]f(x) \le y[/Math] and we know that [Math]\exists x = a[/Math], [Math]\forall x \in E[/Math]. This implies that [Math]f(a) \le y[/Math]
Case 2: Since, [Math]E \subset D[/Math], let [Math]a \notin E[/Math]: We want to show that [Math]f(a) = y[/Math]
I some how need to prove that the cluster point, [Math]a[/Math], lies on the edge of the domain and when plugged in is essentially the real number that bounds [Math]f(x)[/Math]... not sure.

 

[G-X]

Case 2 - Rest:
We know that [Math]f[/Math] is continuous at [Math]a[/Math], by definition, [Math]\forall z \in B_{\delta}(a) \cap D[/Math] s.t. [Math]|f(z) − f(a)| < \epsilon[/Math]. Since [Math]a[/Math] is a cluster point of [Math]E[/Math], by definition, every neighborhood of [Math]a[/Math] contains a point [Math]x \neq a[/Math] where [Math]x \in E[/Math].

Since [Math]E \subset D[/Math], this implies that [Math]\exists z = x \in B_{\delta}(a) \cap D[/Math]. This means that there [Math]\exists x[/Math] s.t. [Math]|f(x) - f(a)| < \epsilon[/Math]
Since [Math]\epsilon > 0[/Math] this means that [Math]0 \ge |f(x) - f(a)|[/Math]. We arrive at an answer when [Math]0 = |f(x) - f(a)|[/Math] or [Math]f(x) = f(a)[/Math]. Since [Math]f(x) \le y[/Math], this implies that [Math]f(a) \le y[/Math].

 

[pka]

Let [Math]f : D[/Math]->[Math] R[/Math] where [Math]D \subset R^{k}[/Math], [Math]k = 1[/Math] or [Math]2[/Math]. Let f be continuous at [Math]a \in D[/Math]and let [Math]a[/Math] be a cluster point of a subset [Math]E[/Math] of [Math]D[/Math]. If there is a real number [Math]y[/Math] such that for each [Math]x \in E,[/Math] [Math]f(x) \le y[/Math], then [Math]f(a) \le y[/Math].

First, what you did in case II is not true.
Do not loose sight of what we want: \(f(a)\le y\)
What if \(f(a)>y\) then if \(\varepsilon=\dfrac{f(a)-y}{2}\).
Can you use continuity to get a contradiction?

 

[G-X]

[Math] |f(x) - f(a)| < \epsilon [/Math] is equivalent to [Math] -\epsilon < f(x) - f(a) < \epsilon[/Math]
So you are saying:

[math]\dfrac{-f(a) + y}{2} < f(x) - f(a)< \dfrac{f(a) - y}{2}[/math]... or [math]- \epsilon < f(x) - f(a)< \dfrac{f(a) - y}{2}[/math] Now I solve this to get a contradiction?

I don't envision f(a) > y... Since a is a cluster point it must reside right next to all the other points in E which map to f(x)? Then can you give me an example of what that would look like? Maybe you already did and I cannot follow yet...

 

[pka]

[Math] |f(x) - f(a)| < \epsilon [/Math] is equivalent to [Math] -\epsilon < f(x) - f(a) < \epsilon[/Math]So you are saying:
[math]\dfrac{-f(a) + y}{2} < f(x) - f(a)< \dfrac{f(a) - y}{2}[/math]... or [math]- \epsilon < f(x) - f(a)< \dfrac{f(a) - y}{2}[/math] Now I solve this to get a contradiction?
I don't envision f(a) > y... Since a is a cluster point it must reside right next to all the other points in E which map to f(x)?

But don't you see that is only true that \(\bf\large\varepsilon= \dfrac{f(a) - y}{2}>0 \) only if \(\bf f(a)>y\)

 

[G-X]

Alright, I see that now. Indeed the definition requires [Math]\epsilon > 0[/Math], but you have chosen a value for f(a) that might not exist? Here you are making f(a) > y... how do you know that exists? That is something we are attempting to prove... Here maybe you have just chosen a negative value...

or rather a value that equals zero...

So you proved my case II??

 

[pka]

Let [Math]f : D[/Math]->[Math] R[/Math] where [Math]D \subset R^{k}[/Math], [Math]k = 1[/Math] or [Math]2[/Math]. Let f be continuous at [Math]\color{red}a \in D[/Math]and let [Math]a[/Math] be a cluster point of a subset [Math]E[/Math] of [Math]D[/Math]. If there is a real number [Math]y[/Math] such that for each [Math]x \in E,[/Math] [Math]f(x) \le y[/Math], then [Math]f(a) \le y[/Math].

Alright, I see that now. Indeed the definition requires [Math]\epsilon > 0[/Math], but you have chosen a value for f(a) that might not exist? Here you are making f(a) > y... how do you know that exists? That is something we are attempting to prove... Here maybe you have just chosen a negative value...

Read the OP. How can \(f(a)\) not exist? Do you understand this question?
I gave you a way to show \(f(a)\le y\) by assuming it is not, i.e. proof by contradiction .

 

[G-X]

To show that f(a) > y to me is not solving the problem in the manner that you did it. The epsilon you have chosen might not exist - prove to me that your chosen epsilon exists? You have created a statement about epsilon, but that does not mean that f(a) can be greater than y. You are supplying a statement you wish to be true.

I understand the definition of continuity and that all epsilon > 0 the statement must hold. So you are setting it equal to a value and attempting to solve for it. Here you have just supplied it a negative epsilon.

[Math]\epsilon = \dfrac{f(x) - y}{2} > 0[/Math]. This must mean then that f(x) > y. But this is false.

 

[G-X]

Since [Math]\epsilon > 0 [/Math], suppose that [Math]f(a) > y[/Math] and let [Math]\epsilon = f(a) - y[/Math]. By substitution, [Math]-f(a) + y < f(x) - f(a) < f(a) - y[/Math]. We can rearrange the equality such that [Math]y < f(x) < 2f(a) - y[/Math] which is a contradiction as [Math]f(x) \le y[/Math]. We can conclude that [Math]f(a) \le y[/Math].