Is $A$ compact and connected

Question

Let $\Bbb R$ be considered with usual topology.

Consider $A=\{x\in \Bbb R:\text{integral part of $x$ is even}\}$ .Show that $A$ is not connected and not compact.

EDITS: $A$ is disconnected as $A=((-\infty ,1)\cap A)\cup ((1,\infty )\cap A)$.

$A$ is unbounded since the set of even integers is a subset of $A$. So $A$ is not compact.

Is it correct?

Answer 1

RE:

Will you please give some hints? Please don't post a full solution.

If you are discussing $\mathbb{R}$ with the "usual topology," then recall that being compact is equivalent to being closed and bounded. Is your set $A$, for example, bounded? (Why or why not?)

To show $A$ is not connected, the typical route is to find a separation using disjoint open sets $S, T \subset \mathbb{R}$ where $S \cup T = A$. Can you find such a separation?