Need help formalizing "T is an open set"

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To show that the set T is open, it is essential to demonstrate that for any point x in T, there exists a neighborhood around x that is entirely contained within T. The discussion emphasizes that if x belongs to T, then there exists a point σ in S such that x is within ε of σ. By defining δ appropriately, it can be shown that any point y within the interval (x - δ, x + δ) will also satisfy the condition for being in T. The use of the triangle inequality is questioned, suggesting that simplifying the approach may clarify the proof. Ultimately, the goal is to establish that T contains an interval around each of its points, confirming its openness.
Terrell
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Homework Statement


Let ##S\subseteq \Bbb{R}## and ##T = \{ t\in \Bbb{R} : \exists s\in S, \vert t-s\vert \lt \epsilon\}## where ##\epsilon## is fixed. I need to show T is an open set.

Homework Equations


n/a

The Attempt at a Solution


Let ##x \in T##, then ##\exists \sigma \in S## such that ##x \in (\sigma -\epsilon, \sigma +\epsilon)##. Let ##\delta = \min(\frac{\vert \sigma -\epsilon -x\vert}{2}),\frac{\vert \sigma +\epsilon -x\vert}{2}## and consider the interval ##(x-\delta, x+\delta)##. If ##y \in (x-\delta, x+\delta)##, then
\begin{align}
\vert \sigma - y \vert \leq \vert \sigma -x\vert + \vert x- y\vert \lt \epsilon + \delta \cdots
\end{align}
I want to show ##\vert \sigma -y \vert \lt \epsilon##, but I am having trouble with finding the right expression to simplify this :cry:
 
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Terrell said:

Homework Statement


Let ##S\subseteq \Bbb{R}## and ##T = \{ t\in \Bbb{R} : \exists s\in S, \vert t-s\vert \lt \epsilon\}## where ##\epsilon## is fixed. I need to show T is an open set.

Homework Equations


n/a

The Attempt at a Solution


Let ##x \in T##, then ##\exists \sigma \in S## such that ##x \in (\sigma -\epsilon, \sigma +\epsilon)##. Let ##\delta = \min(\frac{\vert \sigma -\epsilon -x\vert}{2}),\frac{\vert \sigma +\epsilon -x\vert}{2}## and consider the interval ##(x-\delta, x+\delta)##. If ##y \in (x-\delta, x+\delta)##, then
\begin{align}
\vert \sigma - y \vert \leq \vert \sigma -x\vert + \vert x- y\vert \lt \epsilon + \delta \cdots
\end{align}
I want to show ##\vert \sigma -y \vert \lt \epsilon##, but I am having trouble with finding the right expression to simplify this :cry:
IMO you're using too many symbols: x, y, ##\sigma, \epsilon, \delta##. Let's say you have an interval T of radius r, centered at P: (P - r, P + r). For any point x in this interval, there is some ##\epsilon## so that every point in the interval ##(x - \epsilon, x + \epsilon)## is also in the interval T. The same would not be true of closed intervals.
 
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Mark44 said:
For any point x in this interval, there is some ϵϵ\epsilon so that every point in the interval (x−ϵ,x+ϵ)(x−ϵ,x+ϵ)(x - \epsilon, x + \epsilon) is also in the interval T.
Am I trying to do too much by using the triangle inequality?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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