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2 votes
1 answer
84 views

Find an example of connected topology for finite space $\{1,...,n\}$ that conforms with the intuition of a commoner.

The most common topology for finite space is the discrete topology, which is clearly not connected. There are some other discussion on this site, which give many example topology of finite space and $\...
dodo's user avatar
  • 862
3 votes
2 answers
226 views

Example of a Hausdorff topology that is homeomorphic to $\mathbb R^n$ but isn't second countable?

The definition of a topological manifold $M$ I have is: $M$ is Hausdorff. Each point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$. $M$ is second countable. ...
Makogan's user avatar
  • 3,599
5 votes
3 answers
3k views

Compactness gives Second countable space??

Suppose that $(X,\mathcal{T})$ is a topological space. If we know that $X$ is compact can we assume that is also second countable ?? Because X is compact we have that $X=\cup_{i=1}^{n}V_{i}$, where $...
Jonathan1234's user avatar
  • 1,103
1 vote
1 answer
290 views

Give a subspace of $\mathbb R_l \times \mathbb R_l$ that has no countable dense subset [relative to the subspace topology].

Let $\mathbb R_l$ be $\mathbb R$ given the lower limit topology, i.e. the topology generated by $\{[a, b) \subseteq \mathbb R|a<b\}$. Give a subspace of $\mathbb R_l \times \mathbb R_l$ that has ...
FlickerBeat's user avatar
6 votes
2 answers
3k views

First countable + separable imply second countable? [closed]

In topological space, does first countable+ separable imply second countable? If not, any counterexample?
89085731's user avatar
  • 7,744
6 votes
2 answers
2k views

Why must a locally compact second countable Hausdorff space be second countable to imply paracompactness?

The textbook version of the result I've seen states: A locally compact second countable Hausdorff space is paracompact. Is the property of being second countable needed, or have I missed something? ...
CuriousGeorge's user avatar
14 votes
2 answers
6k views

When is the quotient space of a second countable space second countable?

I am a bit confused about this concept because I have read that the quotient space is second countable if the quotient map is open. However, I thought the definition of a quotient map was a surjective,...
user190570's user avatar
2 votes
1 answer
634 views

If a topological space $S$ is second-countable, must necessarily every quotient space of $S$ be second-countable?

Let $S$ be a second countable topological space. Let $S^*$ be a quotient space of $S$ with quotient map $\pi$. If $\pi$ is open, it's easy to show that it transfers a basis of $S$ into a basis of $S^*$...
ante.ceperic's user avatar
  • 5,311
7 votes
2 answers
2k views

Countable basis but uncountably many connected components

Looking for some guidance on two topology questions: (a) Show that a locally connected space with a countable basis, has at most countably many connected components. (b) Give an example when X has ...
PistolsAtDawn's user avatar
11 votes
2 answers
2k views

Compact topological space not having Countable Basis?

Does there exist a compact topological space not having countable basis? I have constructed a product space from uncountably many unit intervals $[0,1]$, endowed with the product topology. Tychonoff'...
Dilemian's user avatar
  • 1,057
5 votes
2 answers
2k views

locally compact Hausdorff space which is not second-countable

I'm trying to find an example of a space that is Hausdorff and locally compact that is not second countable, but I'm stuck. I search an example on the book Counterexamples in Topology, but I can't ...
User43029's user avatar
  • 1,346
3 votes
2 answers
2k views

Does there exist a Connected Locally Euclidean Space that is not second countable?

A problem in Lee's Introduction to Topological Manifolds got me thinking about this question. I can easily construct a locally euclidean space that is not second countable, by taking a disjoint union ...
JSchlather's user avatar
  • 15.6k