Euclidean and arrow topology

Question

I wonder if the set $A = \{0\} \cup \{\frac1 n; n \in N_1\}$ in the Euclidean topology is compact? If it is compact in the arrow topology, is it compact in Euclidean? It seems to me that it does. Do I understand correctly? Thank you for the hint. $\\:)$

Answer 1

If $X$ is any Hausdorff space and $(x_n)$ is a convergent sequence then $\{x_n\}\cup\{\lim x_n\}$ is a compact space. And in fact it is homeomorphic to your $A$. To see that $A$ is compact you can use the fact that every sequence in $A$ has a convergent subsequence which is relly straight forward. That answers your first question.

As for the arrow topology note that the arrow topology induces the same topology on $A$ as the Euclidean topology. Each $\frac{1}{n}$ is obviously isolated and neighbourhoods of $0$ coincide. Therefore $A$ is the same under Euclidean and arrow topologies.

Answer 2

If for Euclidean topology you mean the topology induced on this set as subspace of the real interval $[0,1]$, then yes it is compact. This follows from the basic fact that a closed subspace of a compact space is compact.

I do not know what is the arrow topology: are you taking as base the sets $[a, +\infty)$ and $(- \infty, b]$?