1
$\begingroup$

I'm having trouble looking for a guideline on how to prove a set is open/closed:

  1. Show that $A = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y + 2x = 3\}$ is closed by showing that every limit point of A is in A.

  2. a) Let $S = \{x \in \mathbb{R} \mid x \not\in \mathbb{Q}\}$, is S closed?

b) Show that $S = \{(x,y) \in \mathbb{R}^{2} \mid xy > 0\}$ is open

c) Let $A,B \subset \mathbb{R}$ with $A$ open, and defined $AB = \{xy \mid (x \in A)\wedge(y \in B)\}$, is $AB$ necessarily open?

Please help me. Thanks!

  • 0
    For example, for (1), showing that "A is closed" really means that you want to show A is a closed subset of R^2. So consider a sequence (x_n) of members of A. These members are also members of the bigger space R^2. If the sequence x_n tends to some limit point x in R as n tends to infinity, then you need to show that x is a member of the set A. That will prove that A is closed (it is a definition of "closed").2012-11-02
  • 0
    Thanks, but can you elaborate further? By looking at the graph, as n->infinity, (x,y)_n -> infinity as well. How do I prove infinity is a member of A?2012-11-02
  • 0
    There isn't really a "general way" since each topology has its own set of quirks. All the topologies you are dealing with here are in Euclidean space, which usually means you need to apply properties about the real numbers and the definition of "open" and "closed" to prove whether a set is open or closed.2012-11-02
  • 0
    Infinity isn't in $A$ - there is no point called infinity in $\mathbb R^2$, so there isn't such a point in $A$.2012-11-02

2 Answers 2