Skip to content
Part CCSIR NET December 2023projection-is-open-map

Projection is open map

Let p:R2→Rp : \mathbb{R}^{2} \to \mathbb{R} be the function p(x, y) = x. Which of the following statements are true?

  1. A.Let A1=A_{1} = {x2+y2<1x^{2} + y^{2} < 1}. For each γ∈p(A1)\gamma \in p(A_{1}) there exists ε>0\varepsilon > 0 with (γ−ε,γ+ε)⊆p(A1)(\gamma - \varepsilon, \gamma + \varepsilon) \subseteq p(A_{1}).✓ Correct
  2. B.Let A2=A_{2} = {x2+y2≤1x^{2} + y^{2} \le 1}. For each γ∈p(A2)\gamma \in p(A_{2}) there exists ε>0\varepsilon > 0 with (γ−ε,γ+ε)⊆p(A2)(\gamma - \varepsilon, \gamma + \varepsilon) \subseteq p(A_{2}).
  3. C.Let A3=A_{3} = {xy = 0}. For each γ∈p(A3)\gamma \in p(A_{3}) there exists ε>0\varepsilon > 0 with (γ−ε,γ+ε)⊆p(A3)(\gamma - \varepsilon, \gamma + \varepsilon) \subseteq p(A_{3}).✓ Correct
  4. D.Let A4=A_{4} = {xy = 1}. For each γ∈p(A4)\gamma \in p(A_{4}) there exists ε>0\varepsilon > 0 with (γ−ε,γ+ε)⊆p(A4)(\gamma - \varepsilon, \gamma + \varepsilon) \subseteq p(A_{4}).✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests finite-dimensional intuition.

See pricing

50 are analysed free — try those first.

The trap it tests

Finite-dimensional intuition

Something true in ℝⁿ, or in a nice space, assumed in general.

Drill statements like this

Related counterexample: If all partial derivatives exist at a point then f is continuous there

More on this topic

The chapter behind this: Differentiability in several variables — free to read

From Functions of Several Variables › Partial derivatives, differentiability, chain rule

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram