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Is the set $U(n,\mathbb R)$ of all upper triangular $n\times n$ matrices over $\mathbb R$ a connected set in $M(n,\mathbb R)$ (with its usual topology after identification with $R^{n^2})?$

I think the answer is yes since connectedness is a productive property, $\mathbb R,\{0\}$ are connected and $$U(n,\mathbb R)=\\\mathbb R\times\mathbb R\times...\times\mathbb R\\\times\{0\}\times \mathbb R\times...\times\mathbb R\\...\\\times\{0\}\times\{0\}\times...\times\mathbb R$$

Please tell me whether the attempt is right or wrong!

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Yes, it is right. Also, you could have constructed by hand the path from one upper triangular matrix to another upper triangular matrix. – Evgeny Aug 9 '13 at 11:04
Also: it is a subspace, so it's even convex. – Daniel Fischer Aug 9 '13 at 11:06

If by "Productive" you mean "stable under product" then your reasoning is right.

You can also show it "by hand": for any two matrices in $U(n,\mathbb R)$, it is not hard to build a continuous path from one to the other.

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There is even a linear segment joining the two matrices, which proves that $U(n,\mathbb R)$ is convex. – lhf Aug 9 '13 at 11:07

It is more, They are path connected too!

According to your notation, $A,B\in U(n,\mathbb{R})$, The continous path from $[0,1]$ is $f(t)=tA+(1-t)B$

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