# Isomorphism and Homeomorphism in Normed Spaces

I have seen a note such as

Let $X$ and $Y$ are normed spaces. If $T:X \to Y$ is a linear isomorphism then it is linear homeomorphism but converse is not ture namely a homeomorphic mapping doesn’t have to be isometry.

Doesn’t linear isomorphism mean $T$ is linear and bijection and linear homeomorphism mean $T$ is linear, bijection, continuous and its inverse is continuous? How can we say being isomorphism requires being homeomorphism? I think its exact opposite is true.

I couldn’t be sure can someone illuminate me please?

Thank you

• It's not true. Perhaps you're just not writing exactly what you saw: If $X$ and $Y$ are complete normed spaces and $T$ is a bounded linear isomorphism then $T$ is a homeomorphism. Jun 11 '18 at 17:11
• @DavidC.Ullrich I know but if they are not complete how can we say if it is linear isomorphism then it is linear homeomorphism? Jun 11 '18 at 17:27
• I already said that that's not true. Jun 11 '18 at 17:34
• @DavidC.Ullrich I’m so sorry. I have misunderstood.I can say if $T$ is a linear homeomorphism then it is isomorphism, can’t I? Jun 11 '18 at 17:36
• Of course not - a homeomorphism need not be linear. Jun 11 '18 at 17:38

To sum up @David C. Ullrich's comments with some examples:

• If $T$ is a linear isomorphism, it doesn't have to be a linear homeomorphism. Consider these examples:

$T : c_{00} \to c_{00}$ given by $(x_n)_n = (nx_n)_n$. Here $T$ is not even bounded.

$S : c_{00} \to c_{00}$ given by $S(x_n)_n = \left(\frac1nx_n\right)_n$. Here $S$ is bounded, but its inverse $T$ is not bounded so $S$ is not a homeomorphism.

Another example: recall that $\ell^1$ and $\ell^2$ both have algebraic dimension equal to $\mathfrak c$ so there exists a linear isomorphism $\ell^1 \to \ell^2$, however there isn't a linear homeomorphism between the two spaces: namely, the dual space of $\ell^2$ is separable while the dual space of $\ell^1$ isn't. However, $\ell^1$ and $\ell^2$ are homeomorphic topological spaces.

• If $T$ is a homeomorphism, then $T$ does not have to be a linear isomorphism. Consider $T : \mathbb{R} \to \mathbb{R}$ given by $T(x) = x^3 + x$. Here $T$ is not linear.

• If $T$ is a linear homeomorphism, then $T$ is a linear isomorphism. This is clear since a homeomorphism is bijective by definition.

• If $T$ is a bounded linear isomorphism between Banach spaces, then $T$ is also a linear homeomorphism. This is a consequence of the Bounded Inverse Theorem.

• It is very very beneficial and clear sum thank you so much. I have given +1 Jun 11 '18 at 20:20