How to prove whether S={(x,y)∈R:y&gt;x2} is open with open set definition

Step 1 Hint: z−w2=(x0−y02)+(z−x0)+(y02−w2) &gt;(x0−y02)−|z−x0|−|w−y0||w+y0| ≥(x0−y02)−|z−x0|−|w−y0|(2y0−|w−y0|) &gt;(x0−y02)−ϵ−ϵ(2y0−ϵ) (Note that y0&gt;0).

Step 1 Let f(x,y)=y−x2 Then, your set becomes: S={(x,y)∈R:y&gt;x2}={(x,y)∈R:y−x2&gt;0} ={(x,y)∈R:f−1(0∞)} The set (0,∞) is an open subset of R and since f is continuous, then f−1(0∞) is also open This concludes the proof.

Step 1 Hint: z−w2=(x0−y02)+(z−x0)+(y02−w2) &gt;(x0−y02)−|z−x0|−|w−y0||w+y0| ≥(x0−y02)−|z−x0|−|w−y0|(2y0−|w−y0|) &gt;(x0−y02)−ϵ−ϵ(2y0−ϵ) (Note that y0&gt;0).

Step 1 Let f(x,y)=y−x2 Then, your set becomes: S={(x,y)∈R:y&gt;x2}={(x,y)∈R:y−x2&gt;0} ={(x,y)∈R:f−1(0∞)} The set (0,∞) is an open subset of R and since f is continuous, then f−1(0∞) is also open This concludes the proof.

Expert answers to "How to prove whether S={(x,y)∈R:y>x2} is open with..."

Ernest Ryland

Answered question

2022-01-12

How to prove whether

S = {(x, y) \in R : y > x^{2}}

is open with open set definition

Answer & Explanation

turtletalk75

Beginner2022-01-13Added 29 answers

Step 1
Hint:

z - w^{2} = (x_{0} - y_{0}^{2}) + (z - x_{0}) + (y_{0}^{2} - w^{2})

> (x_{0} - y_{0}^{2}) - | z - x_{0} | - | w - y_{0} | | w + y_{0} |

\geq (x_{0} - y_{0}^{2}) - | z - x_{0} | - | w - y_{0} | (2 y_{0} - | w - y_{0} |)

> (x_{0} - y_{0}^{2}) - ϵ - ϵ (2 y_{0} - ϵ)

(Note that

y_{0} > 0

Elaine Verrett

Beginner2022-01-14Added 41 answers

Step 1
Let

f (x, y) = y - x^{2}

Then, your set becomes:

S = {(x, y) \in R : y > x^{2}} = {(x, y) \in R : y - x^{2} > 0}

= {(x, y) \in R : f^{- 1} (\begin{array}{cc} 0 & \infty \end{array})}

The set

(0, \infty)

is an open subset of

R

and since f is continuous, then

f^{- 1} (\begin{array}{cc} 0 & \infty \end{array})

is also open
This concludes the proof.

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