As you Know when we define a topological space to be the one point compactification of the topological space $X$, we look for a compact space $Y$ such that $X\subset Y$ and $X$ is dense in $Y$ and $|Y-X|=1$.

But if we add an Extra condition that The space $Y$ to be compact and Hausdorff, We Must eliminate a lot of spaces and Spacial topological spaces Could be in this family that satisfies the **locally compacness** property.

**Theorem:** A topological space $X$ has a Hausdorff one point compactification iff $X$ is locally compact.

But Compared with the above description, I didn't see Any think about the existence of a space with Lindeloff Property except some situation that I Shall describe it as follows:

At first let me define the **lindeloffication** of a topological space:

**Definition:** A Lindeloff space $Y$ is a **one point lindeloffication** of $X$, if $X$ is a dense subspace of $Y$ and $|Y-X|=1$

We Know that any Discrete space $X$ has a **Hausdorff** One point lindelofficatin which defined as follows:

$\rightarrow$ Add a point $p$ to $X$ and consider the set $Y=X\cup${$p$}. then all points of $X$ are open and the Neighborhoods of $p$ is of the form: $U\cup${$p$} where $|X-U|\leq\aleph_0$.

Now my Questions are as follows:

$Q_1$: **For What condition on the space $X$, It has a Hausdorff One point lindeloffication?**

$Q_2$: **Is there an obvious example of space $X$ which is non discrete and has a one point lindeloffication?**

Added Note: When I posed this Question, I didn't notice that It can be occur for a Hausdorff space to Have More than so-called "One point lindeloffication". Gerald Edgar and David Feldman warned to me that this notion is not Functorial.

Its very important to notice that Compactness implies that having one point compactification is functorial or unique up to Homeomorphism. Let me recall the following theorem:

**Theorem1**:For $X$ the following Are equivalent:

- $X$ is maximal compact.
- Every compact subset of $X$ is closed.
- Any continuous bijection $f$ from a compact space $Y$ onto $X$ is a homeomorphism.

For lindelof condition we have the same theorem:

**Theorem2**:For $X$ the following Are equivalent:

**$X$ is maximal Lindelof**.**The set of all closed subsets of $X$ coincides with the set of all Lindelof subspaces of$X$.****If $Y$ is a lindelof space and $f$ is a continuous bijection from $Y$ onto $X$, Then $f$ is a Homeomorphism.**

For the sake of theorem 2 We can find that the one point compactification of an uncountable set is not maximal lindelof.

But We could fix the uniqueness in Question with the maximal lindelofness property as follows:

$Q_3$: **For which topological space, we have a maximal Hausdorff one point lindelofication.**

theone point lindeloffication", but the topology need not be uniquely determined. – Gerald Edgar Jun 17 '12 at 19:02