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.

  • $\begingroup$ You say "the one point lindeloffication", but the topology need not be uniquely determined. $\endgroup$ – Gerald Edgar Jun 17 '12 at 19:02
  • $\begingroup$ Hello Dear Gerald. If The space $X$ has a Hausdorff lindeloffication, Then this space is unique Up to Homeomorphism. the Hausdorff condition is Basic for Uniqueness. $\endgroup$ – Ali Reza Jun 17 '12 at 19:33
  • 4
    $\begingroup$ A discrete space has a one-point lindeloffication as you describe here, but it also has its usual one-point compactification. These are not homeomorphic, are they? $\endgroup$ – Gerald Edgar Jun 17 '12 at 21:58
  • $\begingroup$ Evidently you are right. I didn't notice to it. Thank you very much for your Careful look to my Wrong Description. I corrected it. I have found that where the problem Is. Please notice to added not in the above Question. $\endgroup$ – Ali Reza Jun 18 '12 at 6:50

As for $Q_2$, take $X \times S^1$ with $X$ discrete. The one-point Lindeloffication follows along the same lines as above. Of course any one-point compactification is a fortiori a one-point Lindeloffication, so you have no shortage of examples.

As for $Q_1$, this should help even though theyndon't address your question directly: www.new1.dli.ernet.in/data1/upload/insa/INSA.../20005a7a_876.pdf

  • $\begingroup$ Thank you very much dear David. For $Q_1$ I read the Article which was introduced by You. I think it Answered to my Question Generally. Its very interesting that for having a one point lindeloffication we must increase the condition $T_2$ to condition $T_3$ and consider that if every point has a closed lindeloff nhood! For $Q_2$ I new that all locally compact space has a one point lindeloffication. But I think this Article gives no example of a space which is not locally compact, but has a proper one point lindelofication ? Did you think about the existence of it? $\endgroup$ – Ali Reza Jun 17 '12 at 19:45
  • $\begingroup$ The rationals are Lindelof but not locally compact, and Lindelof spaces almost trivially have one-point Lindelofications. bor But you probably want a non-Lindelof not locally compact space. So take uncountably many copies of the rationals. Make the neighborhoods of the new point contain all but countably many copies. (The one-point compactification is functorial, but I don't see a way of speaking about the one-point Lindelofication.) $\endgroup$ – David Feldman Jun 17 '12 at 21:54

The notion of a one-point Lindeloffication behaves just as well as the notion of the one-point compactification for $P$-spaces. For generality, I am going to answer this question in the context of $P_{\kappa}$-spaces.

A $P$-space is a regular space where the intersection of countably many open sets is open. Every $P$ space is zero-dimensional and hence completely regular. If $\kappa$ is a regular cardinal, then a $P_{\kappa}^{-}$-space is a topological space $X$ where if $U_{i}$ is open for $i\in I$ and $|I|<\kappa$, then $\bigcap_{i\in I}U_{i}$ is open as well, and a $P_{\kappa}$-space is simply a regular $P_{\kappa}^{-}$-space.

If $\lambda$ is a cardinal, then we say that a space $X$ is $\lambda$-compact if every cover of $X$ has a refinement of cardinality less than $\lambda$.

We say that a space $X$ is locally $\lambda$-compact if whenever $U$ is an open set and $x\in U$, there is an open set $V$ with $x\in V\subseteq\overline{V}\subseteq U$ where $\overline{V}$ is $\lambda$-compact.

Using a standard argument, one can show that if $\kappa$ is an uncountable regular cardinal, then every Hausdorff $\kappa$-compact $P_{\kappa}^{-}$-space is regular, normal, paracompact, and even ultraparacompact, and every locally $\kappa$-compact $P_{\kappa}^{-}$-space is zero-dimensional.

If $\kappa$ is a regular cardinal and $(X,\mathcal{T})$ is a Hausdorff locally $\kappa$-compact $P_{\kappa}^{-}$-space, then let $\mathcal{S}$ be the topology on $X\cup\{\infty\}$ where $U\in\mathcal{S}$ if and only if $U\in\mathcal{T}$ or $\infty\in U$ and $U\cap X\in\mathcal{T}$ and $X\setminus U$ is $\kappa$-compact. Then the space $(X\cup\infty,\mathcal{S})$ is a $\kappa$-compact $P_{\kappa}$-space. The one-point $\kappa$-compactification $(X,\mathcal{T})\mapsto(X\cup\infty,\mathcal{S})$ behaves exactly like the one point compactification of a locally compact space. In fact, much of the theory of $P_{\kappa}$-spaces can be obtained simply by replacing every the word “finite” with “less than $\kappa$.” We therefore obtain the one-point Lindeloffication of locally Lindelof $P$-spaces when we set $\kappa=\aleph_{1}.$


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